What is the equation of the tangent to y^2 = 4ax at a point?

The tangent at (x1, y1) on y^2 = 4ax is y*y1 = 2a(x + x1). In slope form: y = mx + a/m with contact point (a/m^2, 2a/m). In parametric form: ty = x + at^2 at the point (at^2, 2at).

Parabola: Chord, Tangent and Normal

Q: What is the equation of the normal to a parabola?

The normal at (x1, y1) on y^2 = 4ax is y - y1 = -(y1/2a)(x - x1). Slope form: y = mx - 2am - am^3 with foot (am^2, -2am). Parametric form: y + tx = 2at + at^3 with foot (at^2, 2at).

Q: Where do the tangents at two points on a parabola meet?

The tangents at parameters t1 and t2 on y^2 = 4ax meet at (a t1 t2, a(t1 + t2)). The y-coordinate is the arithmetic mean of the two ordinates and the x-coordinate is the geometric mean of the two abscissae in magnitude.

Q: What is the reflection property of a parabola?

Any ray parallel to the axis reflects off the parabola through the focus, and any ray from the focus reflects into a ray parallel to the axis. It follows from the tangent bisecting the angle between the focal radius SP and the perpendicular from P onto the directrix. This is used in dish antennas, solar concentrators and headlights.

Q: What is the chord of contact of a parabola?

If the two tangents from an external point (x1, y1) touch y^2 = 4ax at A and B, the chord AB is the chord of contact and its equation is y*y1 = 2a(x + x1).

Q: What is the pair of tangents equation SS1 = T^2?

SS1 = T^2 is the combined equation of the two tangents from an external point (x1, y1) to y^2 = 4ax, where S = y^2 - 4ax, S1 = y1^2 - 4ax1 and T = y*y1 - 2a(x + x1). It has real linear factors only when S1 is positive.

Q: What is the relation t1 t2 = 2 used for?

If the normals at t1 and t2 on y^2 = 4ax meet again on the parabola at t3, then t1 t2 = 2 and t3 = -(t1 + t2). The chord joining t1 and t2 then passes through the fixed point (-2a, 0).

Q: What is the director circle of a parabola?

The director circle is the locus of the intersection of perpendicular tangents. For y^2 = 4ax it degenerates into the line x + a = 0, which is the directrix itself.

Q: Why does ST = SG = SP matter?

Because ST = SP makes triangle STP isosceles, which forces the tangent at P to bisect the angle between the focal radius SP and the perpendicular from P onto the directrix. That angle bisection is the law of reflection at P, which is why parabolic mirrors and antennas focus parallel rays at the focus.

MathsParabolaFor JEE aspirants

A chord of a parabola becomes a tangent when its two ends merge into one point, and the normal at that point is the perpendicular to the tangent there. Almost every result on this page follows from those two sentences. The tangent to $y^{2} = 4 a x$ has three interchangeable forms (point, slope and parametric), the normal has the same three, and from an external point you get two tangents at once through $S S_{1} = T^{2}$ , with their points of contact joined by the chord of contact $y y_{1} = 2 a (x + x_{1})$ . The page closes with the focal properties, above all $S T = S G = S P$ , which is the single fact that explains why a parabolic mirror gathers parallel light at its focus.

Key Formulas - Quick Reference

Chord joining $t_{1}$ and $t_{2}$ : $2 x - (t_{1} + t_{2}) y + 2 a t_{1} t_{2} = 0$ , slope $\frac{2}{t _{1} + t _{2}}$ .
Focal chord condition: $t_{1} t_{2} = - 1$ ; its length is $a (t + \frac{1}{t})^{2}$ .
Tangent at $(x_{1}, y_{1})$ : $y y_{1} = 2 a (x + x_{1})$ .
Tangent with slope $m$ : $y = m x + \frac{a}{m}$ , touching at $(\frac{a}{m ^{2}}, \frac{2 a}{m})$ .
Tangent in parametric form: $t y = x + a t^{2}$ at the point $(a t^{2}, 2 a t)$ .
Conditions of tangency: $y = m x + c$ touches iff $c = \frac{a}{m}$ ; $l x + m y + n = 0$ touches iff $l n = a m^{2}$ .
Intersection of tangents at $t_{1}, t_{2}$ : $(a t_{1} t_{2}, a (t_{1} + t_{2}))$ .
Normal at $(x_{1}, y_{1})$ : $y - y_{1} = - \frac{y _{1}}{2 a} (x - x_{1})$ .
Normal with slope $m$ : $y = m x - 2 am - a m^{3}$ , foot at $(a m^{2}, - 2 am)$ .
Normal in parametric form: $y + t x = 2 a t + a t^{3}$ at the point $(a t^{2}, 2 a t)$ .
Second intersection of a normal with the parabola: if the normal at $t_{1}$ meets the curve again at $t_{2}$ , then $t_{2} = - t_{1} - \frac{2}{t _{1}}$ .
Intersection of normals at $t_{1}, t_{2}$ : $(a (t_{1}^{2} + t_{2}^{2} + t_{1} t_{2} + 2), - a t_{1} t_{2} (t_{1} + t_{2}))$ .
Pair of tangents from $(x_{1}, y_{1})$ : $S S_{1} = T^{2}$ , where $S = y^{2} - 4 a x$ , $S_{1} = y_{1}^{2} - 4 a x_{1}$ , $T = y y_{1} - 2 a (x + x_{1})$ .
Chord of contact from $(x_{1}, y_{1})$ : $y y_{1} = 2 a (x + x_{1})$ .
Chord with midpoint $(x_{1}, y_{1})$ : $T = S_{1}$ , i.e. $y y_{1} - 2 a (x + x_{1}) = y_{1}^{2} - 4 a x_{1}$ .
Director circle: degenerates to the directrix $x + a = 0$ .
Condition for three real normals from $(h, k)$ : $h > 2 a$ and $k^{2} < \frac{4}{27 a} (h - 2 a)^{3}$ .

1. Chords of the Parabola

Everything on this page grows out of one object, so it is worth starting there. Take two points on $y^{2} = 4 a x$ with parameters $t_{1}$ and $t_{2}$ , that is $P (a t_{1}^{2}, 2 a t_{1})$ and $Q (a t_{2}^{2}, 2 a t_{2})$ . The slope of the chord $P Q$ is

\frac{2 a t _{1} - 2 a t _{2}}{a t _{1}^{2} - a t _{2}^{2}} = \frac{2 ( t _{1} - t _{2} )}{( t _{1} - t _{2} ) ( t _{1} + t _{2} )} = \frac{2}{t _{1} + t _{2}},

and the chord itself is

2 x - (t_{1} + t_{2}) y + 2 a t_{1} t_{2} = 0 .

Notice what happens if you let $t_{2} \to t_{1} = t$ : the slope becomes $1/ t$ and the equation collapses to $t y = x + a t^{2}$ . That limiting chord is exactly the tangent at $t$ , which is the cleanest way to see where the parametric tangent comes from.

Chords through a fixed point on the axis

Put $y = 0$ in the chord equation: it meets the axis at $x = - a t_{1} t_{2}$ . So the chord through $t_{1}$ and $t_{2}$ passes through the fixed point $(c, 0)$ on the axis if and only if $t_{1} t_{2} = - \frac{c}{a}$ . Two special cases carry most of the exam weight:

Focal chord (through $S (a, 0)$ ): $t_{1} t_{2} = - 1$ . Its length is $a (t + \frac{1}{t})^{2}$ , which is smallest when $t = \pm 1$ , giving the latus rectum of length $4 a$ .
Chord through $(- 2 a, 0)$ : $t_{1} t_{2} = 2$ . This innocuous-looking case is exactly the condition that turns up later when two normals meet again on the parabola.

2. Tangents to the Parabola $y^{2} = 4 a x$

The tangent at a point can be obtained either by the $T$ -replacement rule (in the equation of the curve substitute $x^{2} \to x x_{1}$ , $y^{2} \to y y_{1}$ , $x \to \frac{x + x _{1}}{2}$ , $y \to \frac{y + y _{1}}{2}$ ) or by differentiating. All three standard forms are worth knowing cold.

Figure 1. The tangent at

P (a t^{2}, 2 a t)

y^{2} = 4 a x

t y = x + a t^{2}

. It meets the curve only at

P

and crosses the axis at

(- a t^{2}, 0)

, the mirror image of the abscissa of

P

Three forms of the tangent

Form	Equation	Point of contact
Point form	$y y_{1} = 2 a (x + x_{1})$	$(x_{1}, y_{1})$
Slope form (with $m \neq = 0$ )	$y = m x + \frac{a}{m}$	$(\frac{a}{m ^{2}}, \frac{2 a}{m})$
Parametric form	$t y = x + a t^{2}$	$(a t^{2}, 2 a t)$

Why slope form excludes $m = 0$ . Setting

m = 0

y = m x + a / m

is meaningless, and geometrically there is a reason: the only tangent to

y^{2} = 4 a x

with zero slope would have to touch at the vertex, but the tangent at the vertex is the vertical line

x = 0

. Every non-vertical tangent has a non-zero slope.

Condition of tangency

A line $y = m x + c$ meets $y^{2} = 4 a x$ where $m^{2} x^{2} + (2 m c - 4 a) x + c^{2} = 0$ . Equal roots require $(2 m c - 4 a)^{2} = 4 m^{2} c^{2}$ , which simplifies to $c = \frac{a}{m}$ . In the general form, $l x + m y + n = 0$ touches $y^{2} = 4 a x$ if and only if $l n = a m^{2}$ .

Tangents to the other standard parabolas

Parabola	Parametric point	Tangent of slope $m$	Normal of slope $m$
$y^{2} = 4 a x$	$(a t^{2}, 2 a t)$	$y = m x + \frac{a}{m}$	$y = m x - 2 am - a m^{3}$
$y^{2} = - 4 a x$	$(- a t^{2}, 2 a t)$	$y = m x - \frac{a}{m}$	$y = m x + 2 am + a m^{3}$
$x^{2} = 4 a y$	$(2 a t, a t^{2})$	$y = m x - a m^{2}$	$x = m y - 2 am - a m^{3}$
$x^{2} = - 4 a y$	$(2 a t, - a t^{2})$	$y = m x + a m^{2}$	$x = m y + 2 am + a m^{3}$

Point of intersection of two tangents

Figure 2. Tangents at

P (t_{1})

and

Q (t_{2})

meet at

T (a t_{1} t_{2}, a (t_{1} + t_{2}))

. The dashed line

P Q

is the chord of contact of

T

The tangents at $t_{1}$ and $t_{2}$ are $t_{1} y = x + a t_{1}^{2}$ and $t_{2} y = x + a t_{2}^{2}$ . Subtracting gives $(t_{1} - t_{2}) y = a (t_{1}^{2} - t_{2}^{2})$ , so $y = a (t_{1} + t_{2})$ , and back-substituting gives $x = a t_{1} t_{2}$ . Hence

T = (a t_{1} t_{2}, a (t_{1} + t_{2})) .

Both coordinates read off neatly. The ordinates of the contact points are $2 a t_{1}$ and $2 a t_{2}$ , whose arithmetic mean is $a (t_{1} + t_{2})$ , exactly the $y$ -coordinate of $T$ . The abscissae are $a t_{1}^{2}$ and $a t_{2}^{2}$ , whose geometric mean is $a ∣ t_{1} t_{2} ∣$ , so the $x$ -coordinate of $T$ is that geometric mean up to sign. In short: $y$ is the AM of the ordinates, $x$ is the GM of the abscissae in magnitude.

3. Normals to the Parabola $y^{2} = 4 a x$

The normal at a point $P$ is the line through $P$ perpendicular to the tangent at $P$ . Since the slope of the tangent at $(x_{1}, y_{1})$ is $\frac{2 a}{y _{1}}$ , the slope of the normal there is $- \frac{y _{1}}{2 a}$ , and at the parametric point it is simply $- t$ .

Figure 3. The normal at

P

is perpendicular to the tangent there (small square), has equation

y + t x = 2 a t + a t^{3}

, and meets the axis at

(2 a + a t^{2}, 0)

Form	Equation	Foot of the normal
Point form	$y - y_{1} = - \frac{y _{1}}{2 a} (x - x_{1})$	$(x_{1}, y_{1})$
Slope form	$y = m x - 2 am - a m^{3}$	$(a m^{2}, - 2 am)$
Parametric form	$y + t x = 2 a t + a t^{3}$	$(a t^{2}, 2 a t)$

Watch the sign convention. In slope form the foot is

(a m^{2}, - 2 am)

, not

(a m^{2}, 2 am)

. The reason is that the slope of the normal at parameter

t

m = - t

, so the point

(a t^{2}, 2 a t)

becomes

(a m^{2}, - 2 am)

. Mixing up

m

and

t

here is the single most common source of sign errors in normal problems.

Where does the normal at $t_{1}$ meet the parabola again?

The normal at $t_{1}$ has slope $- t_{1}$ . If it cuts the curve again at $t_{2}$ , then that line is also the chord joining $t_{1}$ and $t_{2}$ , whose slope is $\frac{2}{t _{1} + t _{2}}$ . Equating the two slopes,

- t_{1} = \frac{2}{t _{1} + t _{2}} ⟹ t_{2} = - t_{1} - \frac{2}{t _{1}} .

Figure 4. The normal at

t_{1}

cuts the parabola a second time at

t_{2} = - t_{1} - \frac{2}{t _{1}}

. Drawn for

t_{1} = 1

, which gives

t_{2} = - 3

So every normal cuts the parabola again at one uniquely determined second point. This is a workhorse identity in JEE Advanced problems on normal chords. Note also that $∣ t_{2} ∣ > ∣ t_{1} ∣$ always, so the second point is further from the vertex than the first.

Point of intersection of two normals

Normals at $t_{1}$ and $t_{2}$ intersect at

(a (t_{1}^{2} + t_{2}^{2} + t_{1} t_{2} + 2), - a t_{1} t_{2} (t_{1} + t_{2})) .

JEE Advanced Two normals meeting on the parabola. If the normals at

t_{1}

and

t_{2}

meet again on the parabola at the point

t_{3}

, then

t_{1} t_{2} = 2

t_{3} = - (t_{1} + t_{2})

, and by Section 1 the chord joining

t_{1}

and

t_{2}

therefore passes through the fixed point

(- 2 a, 0)

4. Conormal Points and Three Real Normals

How many normals pass through a given point $P (h, k)$ ? Requiring $y = m x - 2 am - a m^{3}$ to pass through $(h, k)$ gives

a m^{3} + (2 a - h) m + k = 0.

This is a cubic in $m$ , so at most three normals pass through $P$ . Their slopes $m_{1}, m_{2}, m_{3}$ satisfy, by Vieta's formulas,

m_{1} + m_{2} + m_{3} = 0, m_{1} m_{2} + m_{2} m_{3} + m_{3} m_{1} = \frac{2 a - h}{a}, m_{1} m_{2} m_{3} = - \frac{k}{a} .

The three feet of these normals, call them $A, B, C$ , are the conormal points.

Figure 5. Three normals from

N (h, k)

meet the parabola at the conormal points

A

B

C

. Since

m_{1} + m_{2} + m_{3} = 0

the ordinates also sum to zero, so the centroid of

△ A B C

lies on the axis.

Three consequences of $m_{1} + m_{2} + m_{3} = 0$ :

The algebraic sum of the slopes of the three concurrent normals is zero.
The algebraic sum of the ordinates of the three conormal points is zero, since $y_{i} = - 2 a m_{i}$ .
Therefore the centroid of triangle $A B C$ has $y$ -coordinate zero, so it lies on the axis of the parabola.

JEE Advanced Condition for three real and distinct normals from

(h, k)

y^{2} = 4 a x

. Writing the cubic as

m^{3} + p m + q = 0

with

p = \frac{2 a - h}{a}

and

q = \frac{k}{a}

, three distinct real roots need

4 p^{3} + 27 q^{2} < 0

, which reduces to

h > 2 a and k^{2} < \frac{4}{27 a} (h - 2 a)^{3} .

On the boundary

k^{2} = \frac{4}{27 a} (h - 2 a)^{3}

two of the three normals coincide; that semicubical curve is the evolute of the parabola.

5. Pair of Tangents and Chord of Contact

Pair of tangents from an external point: $S S_{1} = T^{2}$

From a point $P (x_{1}, y_{1})$ lying outside the parabola, exactly two tangents can be drawn. Their combined equation is

S S_{1} = T^{2}, S = y^{2} - 4 a x, S_{1} = y_{1}^{2} - 4 a x_{1}, T = y y_{1} - 2 a (x + x_{1}) .

This is a second-degree equation whose two linear factors are the two tangent lines. The sign of $S_{1}$ decides the whole picture:

$S_{1} > 0$ : the point is outside the parabola and two real distinct tangents exist.
$S_{1} = 0$ : the point is on the parabola and the two tangents coincide.
$S_{1} < 0$ : the point is inside and no real tangent can be drawn.

Chord of contact

Figure 6. From the external point

P (x_{1}, y_{1})

the two tangents touch at

A

and

B

. Together they are the pair of tangents

S S_{1} = T^{2}

; the chord

A B

is the chord of contact

y y_{1} = 2 a (x + x_{1})

When the two tangents from $P (x_{1}, y_{1})$ touch the parabola at $A$ and $B$ , the chord $A B$ is the chord of contact, and its equation is

y y_{1} = 2 a (x + x_{1}) .

This is the same expression as the point-form tangent, but the roles differ completely: there $(x_{1}, y_{1})$ lay on the curve, here it lies outside it.

JEE Advanced Area of the triangle formed by the two tangents from

P (x_{1}, y_{1})

and their chord of contact:

Area = \frac{( y _{1}^{2} - 4 a x _{1} ) ^{3/2}}{2 a} = \frac{S _{1}^{3/2}}{2 a} .

The formula only makes sense when

S_{1} > 0

, which is precisely the condition for the two tangents to exist.

Director circle

The locus of the point of intersection of two perpendicular tangents is called the director circle. Tangents of slopes $m$ and $- 1/ m$ are $y = m x + \frac{a}{m}$ and $y = - \frac{x}{m} - am$ . Eliminating $m$ between them gives $x = - a$ , so the locus is the line

x + a = 0,

which is the directrix itself. The parabola is the odd one out among the conics here: for an ellipse or a hyperbola the director circle is a genuine circle, but for a parabola it degenerates into a straight line.

Chord with a given midpoint: $T = S_{1}$

The chord of $y^{2} = 4 a x$ whose midpoint is $(x_{1}, y_{1})$ has equation

y y_{1} - 2 a (x + x_{1}) = y_{1}^{2} - 4 a x_{1}, that is T = S_{1} .

Its slope is $\frac{2 a}{y _{1}}$ , which depends only on the ordinate of the midpoint. That single observation solves a surprising number of locus problems.

6. Important Focal Properties

These ten properties recur constantly in JEE problems, often as "prove that" questions or as one step inside a longer problem. Throughout, $S (a, 0)$ is the focus.

(i) $S T = S G = S P$ , and the reflection property

If the tangent and the normal at $P$ meet the axis at $T$ and $G$ respectively, then $S T = S G = S P$ . The proof is a one-liner in parametric coordinates: the tangent at $t$ meets $y = 0$ at $T (- a t^{2}, 0)$ and the normal meets it at $G (2 a + a t^{2}, 0)$ , so

S T = a + a t^{2}, S G = a + a t^{2}, S P = a t^{2} + a,

the last using the focal distance $S P = x_{P} + a$ .

Figure 7. The tangent at

P

meets the axis at

T

and the normal at

G

. The tick marks show

S T = S G = S P

, and the two arcs marked

θ

show the tangent bisecting the angle between

P S

and

P M

Because $S T = S P$ , triangle $S T P$ is isosceles, so the tangent makes equal angles with $S P$ and with the axis. Since $P M$ is parallel to the axis, the tangent therefore bisects the angle $S P M$ between the focal radius and the perpendicular onto the directrix. That is the whole content of the reflection property.

Reflection property. Any ray parallel to the axis, on striking the parabola, reflects through the focus; conversely any ray from the focus reflects into a ray parallel to the axis. This is why parabolic mirrors concentrate sunlight, why dish antennas gather signals at a single receiver, and why a headlight with its bulb at the focus throws a parallel beam.

Figure 8. Rays arriving parallel to the axis all reflect through the focus

S

. On the top ray the normal is dashed and the arcs mark the equal angles of incidence and reflection.

(ii) Right angle subtended at the focus

The portion of a tangent cut off between the directrix and the point of contact subtends a right angle at the focus.

(iii) Tangents at the ends of a focal chord

Figure 9. For a focal chord

A B

we have

t_{1} t_{2} = - 1

, so the tangents at

A

and

B

meet at right angles at a point

R

on the directrix. The dashed circle on

A B

as diameter touches the directrix.

If $t_{1}$ and $t_{2}$ are the ends of a focal chord then $t_{1} t_{2} = - 1$ . The tangents there meet at $(a t_{1} t_{2}, a (t_{1} + t_{2})) = (- a, a (t_{1} + t_{2}))$ , a point on the directrix. Their slopes are $\frac{1}{t _{1}}$ and $\frac{1}{t _{2}}$ , whose product is $\frac{1}{t _{1} t _{2}} = - 1$ , so they are perpendicular. Two corollaries follow immediately:

The circle drawn on any focal chord as diameter touches the directrix.
The circle drawn on any focal radius as diameter touches the tangent at the vertex.

(iv) Foot of the perpendicular from the focus onto a tangent

The foot of the perpendicular from $S$ onto any tangent lies on the tangent at the vertex, which for $y^{2} = 4 a x$ is the $y$ -axis.

(v) Image of the focus in any tangent

Reflecting $S$ in any tangent gives a point on the directrix. This is just (iv) restated: the foot of the perpendicular is the midpoint of $S$ and its image, and doubling the distance from the tangent at the vertex $(x = 0)$ lands you on the directrix $(x = - a)$ .

(vi) $S T^{2} = S P \cdot S Q$

If the tangents at $P$ and $Q$ meet at $T$ , then $T P$ and $T Q$ subtend equal angles at $S$ , triangles $S P T$ and $S T Q$ are similar, and

S T^{2} = S P \cdot S Q .

(vii) The semi-latus rectum is the harmonic mean of the focal chord segments

If $P S Q$ is any focal chord, then

\frac{2 \cdot P S \cdot S Q}{P S + S Q} = 2 a, equivalently \frac{1}{P S} + \frac{1}{S Q} = \frac{1}{a} .

(viii) Area ratio for inscribed and circumscribed triangles

The triangle formed by three points on the parabola has twice the area of the triangle formed by the tangents at those same three points.

(ix) and (x) Subtangent and subnormal

Figure 10. The subtangent

T D = 2 \cdot O D

varies with the position of

P

and is bisected at the vertex, while the subnormal

D N = 2 a

is the same for every point of the parabola.

Let $P$ lie on $y^{2} = 4 a x$ , let the tangent and normal at $P$ meet the axis at $T$ and $N$ , and drop a perpendicular from $P$ to the axis meeting it at $D$ . Using $T (- a t^{2}, 0)$ , $D (a t^{2}, 0)$ and $N (2 a + a t^{2}, 0)$ :

Subtangent $T D = 2 a t^{2} = 2 \cdot O D$ , twice the abscissa of $P$ . Equivalently, the vertex $O$ is the midpoint of $T D$ : the subtangent is bisected at the vertex.
Subnormal $D N = 2 a$ , the same constant for every point of the parabola, equal to the semi-latus rectum.

Solved Examples

Solved Example 1

Prove that the line

y = m x + c

touches the parabola

y^{2} = 4 a (x + a)

c = ma + \frac{a}{m}

Solution

Shift the origin with $X = x + a$ , so the curve becomes $Y^{2} = 4 a X$ with $Y = y$ . Its tangent of slope $m$ is $Y = m X + \frac{a}{m}$ . Translating back,

y = m (x + a) + \frac{a}{m} = m x + (ma + \frac{a}{m}) .

Comparing with $y = m x + c$ gives $c = ma + \frac{a}{m}$ . $□$

Solved Example 2

A tangent to the parabola

y^{2} = 8 x

makes an angle of

4 5^{\circ}

with the line

y = 3 x + 5

. Find its equation and its point of contact.

Solution

If the tangent has slope $m$ , then $tan 4 5^{\circ} = \frac{m - 3}{1 + 3 m} = 1$ , so $m - 3 = \pm (1 + 3 m)$ , giving

m (1 \mp 3) = 3 \pm 1 ⟹ m = - 2 or m = \frac{1}{2} .

For $y^{2} = 8 x$ we have $4 a = 8$ , so $a = 2$ . The tangent of slope $m$ is $y = m x + \frac{2}{m}$ , touching at $(\frac{2}{m ^{2}}, \frac{4}{m})$ .

$m = - 2$ : tangent $y = - 2 x - 1$ , contact point $(\frac{1}{2}, - 2)$ .
$m = \frac{1}{2}$ : tangent $y = \frac{1}{2} x + 4$ , contact point $(8, 8)$ .

Solved Example 3

Find the equations of the tangents to

y^{2} = 9 x

passing through the point

(4, 10)

Solution

Here $4 a = 9$ , so $a = \frac{9}{4}$ and a tangent of slope $m$ is $y = m x + \frac{9}{4 m}$ . Requiring it to pass through $(4, 10)$ ,

10 = 4 m + \frac{9}{4 m} ⟹ 16 m^{2} - 40 m + 9 = 0 ⟹ m = \frac{1}{4} or \frac{9}{4} .

The two tangents are

y = \frac{x}{4} + 9 and y = \frac{9 x}{4} + 1.

Two real tangents were guaranteed in advance, since $S_{1} = 1 0^{2} - 9 (4) = 64 > 0$ places $(4, 10)$ outside the parabola.

Solved Example 4

Find the common tangent of the parabolas

y^{2} = 4 a x

and

x^{2} = 4 b y

Solution

Any tangent to $y^{2} = 4 a x$ is $y = m x + \frac{a}{m}$ . Substituting into $x^{2} = 4 b y$ ,

x^{2} = 4 b (m x + \frac{a}{m}) ⟹ x^{2} - 4 bm x - \frac{4 ab}{m} = 0.

For this line to touch the second parabola as well, the roots must be equal, so the discriminant vanishes:

16 b^{2} m^{2} + \frac{16 ab}{m} = 0 ⟹ b m^{3} = - a ⟹ m = - (\frac{a}{b})^{1/3} .

Then the intercept is $\frac{a}{m} = - a (\frac{b}{a})^{1/3} = - a^{2/3} b^{1/3}$ , so the common tangent is

y = - (\frac{a}{b})^{1/3} x - a^{2/3} b^{1/3}, or a^{1/3} x + b^{1/3} y + a^{2/3} b^{2/3} = 0 .

The cubic $b m^{3} = - a$ has exactly one real root, so the two parabolas have exactly one common tangent.

Solved Example 5

If the normal at

t_{1}

y^{2} = 4 a x

meets the parabola again at

t_{2}

, show that

t_{2} = - t_{1} - \frac{2}{t _{1}}

Solution

The normal at $P (t_{1}) = (a t_{1}^{2}, 2 a t_{1})$ has slope $- t_{1}$ . Since it cuts the curve again at $t_{2}$ , that same line is the chord joining $t_{1}$ and $t_{2}$ , whose slope is $\frac{2}{t _{1} + t _{2}}$ from the chord equation $2 x - (t_{1} + t_{2}) y + 2 a t_{1} t_{2} = 0$ . Therefore

- t_{1} = \frac{2}{t _{1} + t _{2}} ⟹ t_{1} + t_{2} = - \frac{2}{t _{1}} ⟹ t_{2} = - t_{1} - \frac{2}{t _{1}} . □

Solved Example 6

If the normals at

t_{1}

and

t_{2}

meet again on the parabola at

t_{3}

, prove that

t_{1} t_{2} = 2

and

t_{1} + t_{2} + t_{3} = 0

Solution

Apply Example 5 to each normal:

t_{3} = - t_{1} - \frac{2}{t _{1}} (normal at t_{1}), t_{3} = - t_{2} - \frac{2}{t _{2}} (normal at t_{2}) .

Equating the two expressions,

(t_{2} - t_{1}) + 2 (\frac{1}{t _{2}} - \frac{1}{t _{1}}) = 0.

Using $\frac{1}{t _{2}} - \frac{1}{t _{1}} = \frac{t _{1} - t _{2}}{t _{1} t _{2}}$ , this becomes

(t_{2} - t_{1}) (1 - \frac{2}{t _{1} t _{2}}) = 0.

Since $t_{1} \neq = t_{2}$ , we get $t_{1} t_{2} = 2$ . Substituting back, $t_{3} = - t_{1} - \frac{2}{t _{1}} = - t_{1} - t_{2}$ , hence $t_{1} + t_{2} + t_{3} = 0$ . $□$

Solved Example 7

Write the equation of the pair of tangents drawn from

P (- 1, 2)

to the parabola

y^{2} = 4 x

Solution

Here $a = 1$ , $S = y^{2} - 4 x$ , $S_{1} = 2^{2} - 4 (- 1) = 8$ , and $T = 2 y - 2 (x - 1) = 2 y - 2 x + 2$ . Since $S_{1} = 8 > 0$ , the point is external and two tangents exist. The pair of tangents is $S S_{1} = T^{2}$ :

8 (y^{2} - 4 x) = (2 y - 2 x + 2)^{2} = 4 (y - x + 1)^{2} .

Dividing by $4$ and expanding the right side,

2 y^{2} - 8 x = x^{2} + y^{2} + 1 - 2 x y - 2 x + 2 y .

Collecting all terms on one side,

y^{2} - x^{2} + 2 x y - 6 x - 2 y - 1 = 0 .

Sanity check. The point

(- 1, 2)

lies on the directrix

x = - 1

, so by the director circle property the two tangents must be perpendicular. Indeed, in the answer the coefficients of

x^{2}

and

y^{2}

are

- 1

and

+ 1

, and their sum is zero, which is exactly the condition for a pair of perpendicular lines.

Solved Example 8

Find the locus of the point from which the two tangents to

y^{2} = 4 a x

have slopes

m_{1}, m_{2}

with

m_{1} + m_{2} = m_{0}

, a constant.

Solution

Let $P = (h, k)$ . The tangent $y = m x + \frac{a}{m}$ passes through $P$ when $k = mh + \frac{a}{m}$ , that is

h m^{2} - k m + a = 0.

The two roots are the slopes of the two tangents, so $m_{1} + m_{2} = \frac{k}{h}$ . Setting this equal to $m_{0}$ gives $k = m_{0} h$ , and replacing $(h, k)$ by $(x, y)$ ,

y = m_{0} x .

The locus is a straight line through the origin of slope $m_{0}$ , with the vertex itself excluded.

Solved Example 9

The line

x - y - 1 = 0

meets

y^{2} = 8 x

P

and

Q

. Find the point of intersection of the tangents at

P

and

Q

Solution

Let the tangents meet at $T (h, k)$ . Then $P Q$ is the chord of contact of $T$ , and with $a = 2$ that chord is $y k = 4 (x + h)$ , i.e. $4 x - k y + 4 h = 0$ . This must be the same line as $x - y - 1 = 0$ , so comparing coefficients,

\frac{4}{1} = \frac{- k}{- 1} = \frac{4 h}{- 1} ⟹ k = 4, h = - 1.

The tangents meet at $(- 1, 4)$ .

Solved Example 10

Find the locus of the midpoint of a chord of

y^{2} = 4 a x

that passes through the fixed point

(p, q)

Solution

Let $M (h, k)$ be the midpoint. The chord with this midpoint is $T = S_{1}$ :

y k - 2 a (x + h) = k^{2} - 4 ah .

Since it passes through $(p, q)$ , substitute $x = p$ , $y = q$ :

q k - 2 a p - 2 ah = k^{2} - 4 ah ⟹ k^{2} - q k - 2 ah + 2 a p = 0.

Replacing $(h, k)$ by $(x, y)$ , the locus is

y^{2} - 2 a x - q y + 2 a p = 0,

itself a parabola with the same axis direction as the original.

Solved Example 11

Find the locus of the midpoints of chords of

y^{2} = 4 a x

that all have slope

m

Solution

The chord with midpoint $(h, k)$ is $y k - 2 a (x + h) = k^{2} - 4 ah$ , whose slope is $\frac{2 a}{k}$ . Setting $\frac{2 a}{k} = m$ gives $k = \frac{2 a}{m}$ , so the locus is the horizontal line

y = \frac{2 a}{m} .

Every set of parallel chords of a parabola therefore has its midpoints on a line parallel to the axis. That line is called the diameter of the system of parallel chords.

Solved Example 12

Find the locus of the point from which three normals are drawn to

y^{2} = 4 a x

such that two of them are perpendicular to each other.

Solution

Let $N = (h, k)$ . The slopes of the three normals satisfy $a m^{3} + (2 a - h) m + k = 0$ , hence

m_{1} + m_{2} + m_{3} = 0, m_{1} m_{2} + m_{2} m_{3} + m_{3} m_{1} = \frac{2 a - h}{a}, m_{1} m_{2} m_{3} = - \frac{k}{a} .

Two normals are perpendicular, so $m_{1} m_{2} = - 1$ . From the product relation, $- m_{3} = - \frac{k}{a}$ , giving $m_{3} = \frac{k}{a}$ , and from the sum relation $m_{1} + m_{2} = - m_{3} = - \frac{k}{a}$ . Substituting into the middle relation,

m_{1} m_{2} + m_{3} (m_{1} + m_{2}) = - 1 + \frac{k}{a} (- \frac{k}{a}) = \frac{2 a - h}{a} .

Multiplying through by $a^{2}$ : $- a^{2} - k^{2} = a (2 a - h)$ , so $k^{2} = ah - 3 a^{2}$ . Replacing $(h, k)$ by $(x, y)$ ,

y^{2} = a (x - 3 a) .

The locus is another parabola, congruent to the original one but with its vertex shifted to $(3 a, 0)$ .

Common Mistakes to Avoid

Watch out

Wrong sign in the slope form of the normal. It is $y = m x - 2 am - a m^{3}$ , with minus signs on both trailing terms, and the foot is $(a m^{2}, - 2 am)$ . Mixing up $m$ and $t$ here (they differ by a sign, $m = - t$ ) is the commonest slip in normal problems.
Confusing the chord of contact with the pair of tangents. From $(x_{1}, y_{1})$ the chord of contact is the single line $y y_{1} = 2 a (x + x_{1})$ ; the pair of tangents is the second-degree equation $S S_{1} = T^{2}$ . One is a line, the other is two lines.
Applying $t_{1} t_{2} = 2$ to any two normals. That relation holds only when the two normals meet again on the parabola. For three normals concurrent at a general point $(h, k)$ , use the Vieta relations $m_{1} + m_{2} + m_{3} = 0$ , $m_{1} m_{2} + m_{2} m_{3} + m_{3} m_{1} = \frac{2 a - h}{a}$ , $m_{1} m_{2} m_{3} = - \frac{k}{a}$ instead.
Forgetting that $S_{1} > 0$ is required for a pair of tangents. If $(x_{1}, y_{1})$ lies inside the parabola then $S_{1} < 0$ , no real tangents exist, and $S S_{1} = T^{2}$ has no real linear factors. Check the position of the point first.
Using $T = S_{1}$ for a chord through a given point. $T = S_{1}$ gives the chord whose midpoint is $(x_{1}, y_{1})$ , not a chord passing through $(x_{1}, y_{1})$ . These are different chords with different equations.
Assuming the director circle is a genuine circle. For a parabola it degenerates into the directrix $x + a = 0$ . This collapse is unique to the parabola; for ellipses and hyperbolas the director circle really is a circle.
Confusing subtangent with subnormal. The subtangent $T D = 2 \cdot O D$ varies from point to point, while the subnormal $D N = 2 a$ is the same everywhere. They are equal only when $O D = a$ , that is at the ends of the latus rectum.

Frequently Asked Questions

Q1. What is the equation of the tangent to $y^{2} = 4 a x$ at $(x_{1}, y_{1})$ ?

The tangent at $(x_{1}, y_{1})$ is $y y_{1} = 2 a (x + x_{1})$ . It can also be written in slope form as $y = m x + \frac{a}{m}$ , touching at $(\frac{a}{m ^{2}}, \frac{2 a}{m})$ , or in parametric form as $t y = x + a t^{2}$ , touching at $(a t^{2}, 2 a t)$ .

Q2. What is the equation of the normal to a parabola?

The normal at $(x_{1}, y_{1})$ on $y^{2} = 4 a x$ is $y - y_{1} = - \frac{y _{1}}{2 a} (x - x_{1})$ . In slope form it is $y = m x - 2 am - a m^{3}$ with foot at $(a m^{2}, - 2 am)$ , and in parametric form $y + t x = 2 a t + a t^{3}$ with foot at $(a t^{2}, 2 a t)$ .

Q3. Where do the tangents at two points on a parabola meet?

The tangents at $t_{1}$ and $t_{2}$ on $y^{2} = 4 a x$ meet at $(a t_{1} t_{2}, a (t_{1} + t_{2}))$ . The $y$ -coordinate is the arithmetic mean of the two ordinates, and the $x$ -coordinate is the geometric mean of the two abscissae in magnitude.

Q4. What is the reflection property of a parabola?

Any ray parallel to the axis reflects off the parabola through the focus, and conversely any ray from the focus reflects into a ray parallel to the axis. It follows from the tangent bisecting the angle between the focal radius $S P$ and the perpendicular from $P$ onto the directrix. This is the principle behind dish antennas, solar concentrators and headlight reflectors.

Q5. What is the chord of contact of a parabola?

If the two tangents from an external point $(x_{1}, y_{1})$ touch $y^{2} = 4 a x$ at $A$ and $B$ , the line $A B$ is the chord of contact and its equation is $y y_{1} = 2 a (x + x_{1})$ . It is the same expression as the point-form tangent, but here $(x_{1}, y_{1})$ lies outside the curve rather than on it.

Q6. What is the pair of tangents equation $S S_{1} = T^{2}$ ?

It is the combined equation of the two tangents from $(x_{1}, y_{1})$ to $y^{2} = 4 a x$ , where $S = y^{2} - 4 a x$ , $S_{1} = y_{1}^{2} - 4 a x_{1}$ and $T = y y_{1} - 2 a (x + x_{1})$ . It is a second-degree equation whose two linear factors are the tangent lines, and it has real factors only when $S_{1} > 0$ .

Q7. How many normals can be drawn from a point to a parabola?

Substituting a point into the slope form of the normal gives a cubic in $m$ , so at most three normals pass through any point. All three are real and distinct when $h > 2 a$ and $k^{2} < \frac{4}{27 a} (h - 2 a)^{3}$ . Their feet are the conormal points, and since the slopes sum to zero, the centroid of the triangle they form lies on the axis.

Q8. What is the relation $t_{1} t_{2} = 2$ used for?

If the normals at $t_{1}$ and $t_{2}$ meet again on the parabola at a third point $t_{3}$ , then $t_{1} t_{2} = 2$ and $t_{3} = - (t_{1} + t_{2})$ . Because a chord through $t_{1}$ and $t_{2}$ meets the axis at $- a t_{1} t_{2}$ , the chord then passes through the fixed point $(- 2 a, 0)$ . This combination appears often in Advanced-level conormal problems.

Q9. What is the director circle of a parabola?

The director circle is the locus of the intersection of perpendicular tangents. For $y^{2} = 4 a x$ it degenerates into the line $x + a = 0$ , the directrix itself. Unlike the ellipse and hyperbola, whose director circles are genuine circles, the parabola's is a straight line.

Q10. Why does $S T = S G = S P$ matter?

Because $S T = S P$ makes triangle $S T P$ isosceles, which forces the tangent at $P$ to bisect the angle between the focal radius $S P$ and the perpendicular from $P$ onto the directrix. That angle bisection is precisely the law of reflection at $P$ , so a ray arriving parallel to the axis leaves along $P S$ . It is the geometric reason parabolic mirrors and antennas work at all.

Previous year questions on Parabola: Chord, Tangent and Normal

19 questions from past papers, each with a step-by-step solution.

Show all 19 questions

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1. Chords of the Parabola

Chords through a fixed point on the axis

2. Tangents to the Parabola y2=4ax

Three forms of the tangent

Condition of tangency

Tangents to the other standard parabolas

Point of intersection of two tangents

3. Normals to the Parabola y2=4ax

Where does the normal at t1​ meet the parabola again?

Point of intersection of two normals

4. Conormal Points and Three Real Normals

5. Pair of Tangents and Chord of Contact

Pair of tangents from an external point: SS1​=T2

Chord of contact

Director circle

Chord with a given midpoint: T=S1​

6. Important Focal Properties

(i) ST=SG=SP, and the reflection property

(ii) Right angle subtended at the focus

(iii) Tangents at the ends of a focal chord

(iv) Foot of the perpendicular from the focus onto a tangent

(v) Image of the focus in any tangent

(vi) ST2=SP⋅SQ

(vii) The semi-latus rectum is the harmonic mean of the focal chord segments

(viii) Area ratio for inscribed and circumscribed triangles

(ix) and (x) Subtangent and subnormal

Solved Examples

Common Mistakes to Avoid

Frequently Asked Questions

Q1. What is the equation of the tangent to y2=4ax at (x1​,y1​)?

Q2. What is the equation of the normal to a parabola?

Q3. Where do the tangents at two points on a parabola meet?

Q4. What is the reflection property of a parabola?

Q5. What is the chord of contact of a parabola?

Q6. What is the pair of tangents equation SS1​=T2?

Q7. How many normals can be drawn from a point to a parabola?

Q8. What is the relation t1​t2​=2 used for?

Q9. What is the director circle of a parabola?

Q10. Why does ST=SG=SP matter?

Previous year questions on Parabola: Chord, Tangent and Normal

Ready to master Parabola?

2. Tangents to the Parabola $y^{2} = 4 a x$

3. Normals to the Parabola $y^{2} = 4 a x$

Where does the normal at $t_{1}$ meet the parabola again?

Pair of tangents from an external point: $S S_{1} = T^{2}$

Chord with a given midpoint: $T = S_{1}$

(i) $S T = S G = S P$ , and the reflection property

(vi) $S T^{2} = S P \cdot S Q$

Q1. What is the equation of the tangent to $y^{2} = 4 a x$ at $(x_{1}, y_{1})$ ?

Q6. What is the pair of tangents equation $S S_{1} = T^{2}$ ?

Q8. What is the relation $t_{1} t_{2} = 2$ used for?

Q10. Why does $S T = S G = S P$ matter?