Parabola: Chord, Tangent and Normal

Points of Intersection of a Straight Line with the Parabola:

Points of intersection of y² = 4ax and y = mx + c are given by (mx + c)² = 4ax

i.e. m²x² + 2x(mc - 2a) + c² = 0 ……(1)

Since (1) is a quadratic equation, the straight line meets the parabola in two points (real, coincident, or imaginary). The roots of (1) are real or imaginary according as

{2(mc – 2a)}² - 4m²c² is positive or negative, i.e. according as –amc +a² is positive or negative, i.e. according as mc is less than or greater than a.

Note:

$\bullet$When m is very small, one of the roots of equation (1) is very large; when m is equal to zero, this root is infinitely large. Hence every straight line parallel to the axis of the parabola meets the curve in one point at a finite distance and in another point at an infinite distance from the vertex. It means that a line parallel to the axis of the parabola meets the parabola only in one point.

Length of the Chord:

As in the preceding article, the abscissae of the points common to the straight line y = mx + c and the parabola y² = 4ax are given by the equation m² x² + ( 2mc – 4a) x + c² = 0.

If (x₁, y₁) and (x₂, y₂) are the points of intersection, then

(x₁ - x₂)² = (x₁ + x₂)² - 4x₁ x₂

= $\frac{4{\left(mc-2a\right)}^2}{m^4}-\frac{4c^2}{m^2}=\frac{16a\left(a-mc\right)}{m^4}$ and (y₁ - y₂) = m(x₁ - x₂)

Hence, the required length

=$\sqrt{{\left({\text{y}}_{\text{1}}-{\text{y}}_{\text{2}}\right)}^{\text{2}}+{\left({\text{x}}_{\text{1}}-{\text{x}}_{\text{2}}\right)}^{\text{2}}}=\sqrt{\text{1}+{\text{m}}^{\text{2}}}\left|\left({\text{x}}_{\text{1}}-{\text{x}}_{\text{2}}\right)\right|=\frac{4}{m^2}\sqrt{1+m^2}\sqrt{a\left(a-mc\right)}$

TANGENT TO A PARABOLA

Tangent at the Point (x₁, y₁):

Let the equation of the parabola be y² = 4ax.

Hence, value of $\frac{\text{dy}}{\text{dx}}$ at P(x₁, y₁) is $\frac{2a}{y_1}$ and the equation of the tangent at P is

y - y₁ = $\frac{\text{2a}}{{\text{y}}_{\text{1}}}$(x - x₁) i.e. yy₁ = 2a(x - x₁) + y₁² $\Rightarrow$ yy₁ = 2a(x + x₁)

Tangent in Terms of m:

Suppose that the equation of a tangent to the parabola y² = 4ax ……(1)

is y = mx + c ……(2)

The abscissae of the points of intersection of (1) and (2) are given by the equation

(mx + c)² = 4ax. But the condition that the straight line (ii) should touch the parabola is that it should meet the parabola in coincident points

$\Rightarrow$ (mc - 2a)² = m²c² ……(3)

$\Rightarrow$ c = a /m.

Hence, y = mx + a/m is a tangent to the parabola y² = 4ax, whatever be the value of m.

Equation (mx +c)² = 4ax now becomes (mx - a/m)² = 0

$\Rightarrow$ x = $\frac{\text{a}}{{\text{m}}^{\text{2}}}$ and y² = 4ax $\Rightarrow$ y = $\frac{\text{2a}}{\text{m}}$.

Thus the point of contact of the tangent y = mx + a/m is $\left(\frac{a}{m^2},\frac{2a}{m}\right)$.

Tangent at the Point 't':

Let the equation of the parabola is y² = 4ax.

The equation of the tangent at (x₁, y₁) to this parabola is yy₁ = 2a(x + x₁).

If the point (x₁, y₁) º (at², 2at)

Equation of tangent becomes y.2at = 2a(x + at²) $\Rightarrow$ yt = x + at².

Note:

$\bullet$The point of intersection of the tangents at 't₁' and 't₂' to the parabola y² = 4ax is

(at₁t₂, a(t₁ + t₂)).

Example -1: Two tangents are drawn from the point (-2, -1) to the parabola y² = 4x. If $\alpha$ is the angle between these tangents, then tan $\alpha$ equals

(A) 3 (B) 1/3

(C) 2 (D) 1/2

Solution: Here a = 1. Any tangent is y = mx + $\frac{1}{m}$.

It passes through (-2, -1)

$\therefore$ 2m² –m –1 = 0

$\therefore$ m = 1, $-\frac{1}{2}$ $\Rightarrow$ tan q = $\frac{1+\frac{1}{2}}{1-\frac{1}{2}}=3$

Hence (A) is the correct answer.

Example -2: If the line 2x + 3y = 1 touches the parabola y² = 4ax, find the length of the latus rectum.

Solution: Equation of any tangent to y² = 4ax is

$y=mx+\frac{a}{m}$ $\Rightarrow$ m²x – my + a = 0

Comparing it with the given tangent 2x + 3y – 1 = 0, we find

$\frac{m^2}{2}=\frac{-m}{3}=\frac{a}{-1}$ $\Rightarrow$ $m=\frac{-2}{3},a=\frac{m}{3}=-\frac{2}{9}$

Hence the length of the latus rectum = 4a = $\frac{8}{9}$

ignoring the negative sign for length.

Equation of the Tangents from an External Point:

Let y² = 4ax be the equation of a parabola and (x₁, y₁) an external point P. Then, equations of the tangents from (x₁, y₁) to the given parabola are given by

SS₁ = T², where S = y² - 4ax, S₁ = y₁² - 4ax₁, T = yy₁ - 2a(x + x₁)

Chord of Contact:

Equation to the chord of contact of the tangents drawn from a point (x₁, y₁), to the parabola y²= 4ax is T= 0, i.e. yy₁ - 2a(x+x₁) =0.

Equation of a Chord with Midpoint (x₁, y₁):

The equation of the chord of the parabola y²= 4ax with mid point (x₁, y₁) is T= S₁

i.e. yy₁-2a(x+x₁)= y₁²-4ax₁ Or yy₁ - 2ax = y₁² - 2ax₁ .

Example -3: Find the equation of the chord of the parabola y² = 12x which is bisected at the point (5, –7).

Solution: Here (x₁, y₁) = (5 –7) and y² = 12x, $\therefore$a = 3

The equation of the chord is S₁ = T

or $y_1^2$ - 4ax₁ = yy₁ – 2a (x + x₁)

or (–7)² – 12.5 = y(–7) – 6(x+5)

$\Rightarrow$ 6x + 7y + 19 = 0.

NORMAL TO THE PARABOLA

Normal at the Point (x₁, y₁ ):

The equation of the tangent at the point (x₁, y₁) is yy₁ = 2a(x + x₁). Since the slope of

tangent = 2a/y₁ , slope of normal is -y₁/ 2a . Also it passes through (x₁, y₁).

Hence its equation is y - y₁ = $-\frac{y_1}{2a}\left(x-x_1\right)$ . . . . . (i)

Normal in Terms of m:

In equation (i), put $-\frac{{\text{y}}_{\text{1}}}{\text{2a}}=\text{m}$ so that y₁ = -2am and x₁ = $\frac{{{\text{y}}_{\text{1}}}^{\text{2}}}{\text{4a}}=\text{a}{\text{m}}^{\text{2}}$, then the equation becomes y = mx - 2am - am³ . . . . . (ii)

where m is a parameter. Equation (ii) is the normal at the point (am², -2am) of the parabola.

Notes:

$\bullet$ If this normal passes through a point (h, k), then k = mh – 2am - am³.

For a given parabola and a given point (h, k) , this cubic in m has three roots say m₁, m₂, m₃ i.e. from (h, k) three normals can be drawn to the parabola whose slopes are m₁, m₂, m₃ . For the cubic, we have

m₁+ m₂ + m₃ = 0

m₁ m₂ +m₂ m₃ +m₃ m₁ = (2a-h) /a

m₁ m₂ m₃ = - k/a

If we have an extra condition about the normals drawn from a point (h, k) to a given parabola y² =4ax then by eliminating m₁, m₂, m₃ from these four relations between m₁, m₂, m₃, we can get the locus of (h, k).

$\bullet$ Since the sum of the roots is equal to zero, the sum of the ordinates of the feet of the normals from a given point is zero.

Normal at the Point 't':

Equation of the normal to y² = 4ax at the point (x₁, y₁) is y – y₁ = –$\frac{y_1}{2a}$(x – x₁)

If (x₁, y₁) º (at², 2at)

Equation of normal becomes

y – 2at = –$\frac{2at}{2a}$(x – at²) $\Rightarrow$ y = –tx + 2at + at³.

Notes:

$\bullet$ If normal at the point t₁ meets the parabola again at the point t₂, then t₂ = -t₁ –2/t₁.

$\bullet$ Point of intersection of the normals to the parabola y² = 4 ax at (at₁², 2at₁) and

(at₂², 2at₂) is (2a + a(t₁² + t₂² + t₁t₂),– at₁t₂(t₁+ t₂)) .

Example -4: Find the locus of the foot of the perpendicular drawn from the vertex on a tangent to the parabola y² = 4ax.

Solution: Any tangent of slope m to the parabola y² = 4ax is

y = mx + a/m . . . (1)

The perpendicular from the vertex on (1) is

x + my = 0 . . . . (2)

By eliminating m between (1) and (2) we obtain the required locus as xy² + x³ + ay² = 0

Example -5: If the normal at the point (at₁², 2at₁) meets the parabola y² = 4ax again at the point (at₂², 2at₂), prove that t₂ = –$t_1-\frac{2}{t_1}$

Solution: The equation of the normal at (at₁², 2at₁) is

y = –t₁x + 2at₁ + at₁³.

Since the normal passes through the point (at₂², 2at₂),

we have, 2at₂ = -t₁. at₂² +2at₁ + at₁³

$\Rightarrow$ 2a (t₁ – t₂) = at₁ t₂² – at₁³

= at₁ $(t_2^2-t_1^2)$ = at₁(t₂ – t₁) (t₂ + t₁)

$\Rightarrow$ 2 = –t₁ (t₂ + t₁), as t₁ ¹ t₂

$\Rightarrow$ t₂ + t₁ = $-\frac{2}{t_1}$ $\Rightarrow$ t₂ = –t₁ $-\frac{2}{t_1}$

Subtangent and Subnormal:

Let the tangent and normal at any point

P (x₁, y₁) on the parabola y² = 4ax meet the axis in T and G respectively. Then PT is called the length of the tangent at P and PG is called the length of the normal at P.

NT is called the subtangent and NG the subnormal at P. The coordinates of T and G can be easily found by putting x = 0 in the equations of the tangent and normal at P. It is evident from figure that

Subtangent = NT = twice the abscissa of P

Subnormal = NG = 2a = semilatus rectum

Example -6: If P is a point on the parabola y² = 4ax such that the subtangent and subnormal at P are equal, find the coordinates of P.

Solution: Let P (x, y) be the required point.

Length of subtangent = twice the abscisse of P = 2x

Length of subnormal = 2a

Since subtangent = subnormal, 2x = 2a $\Rightarrow$ x = a $\Rightarrow$ y² = 4a² $\Rightarrow$ y = $\pm$ 2a

$\therefore$ The required points are (a, 2a) and (a, –2a).

PROPERTIES OF THE PARABOLA

(i) The tangent at any point P on a parabola bisects the angle between the focal chord through P and the perpendicular from P on the directrix.

In the given figure $\angle$MPT = $\angle$TPS :

Similarly the normal at any point on a parabola bisects the angle between the focal chord and the line parallel to the axis through that point.

(ii) The portion of a tangent to a parabola cut off between the directrix and the curve subtends a right angle at the focus.

In the given figure SP is perpendicular to SK i.e. $\angle$KSP = 90⁰ .

(iii) Tangents at the extremities of any focal chord intersect at right angles on the directrix.

(iv) Any tangent to a parabola and the perpendicular on it from the focus meet on the tangent at the vertex.