Introduction to Parabola

Standard equation of a Parabola:

Let S be the focus, ZM the directrix and P the moving point. Draw SZ perpendicular from S on the directrix. Then SZ is the axis of the parabola. Now the middle point of SZ, say A, will lie on the locus of P, i.e., AS = AZ. Take A as the origin, the x-axis along AS, and the y- axis along the perpendicular to AS at A, as in the figure.

Let AS = a, so that ZA is also a. Let (x, y) be the coordinates of the moving point P. Then MP = ZN = ZA + AN = a + x. But by definition MP = PS $\Rightarrow$ MP² = PS²

So that, (a + x)² = (x – a)² + y².

Hence, the equation of parabola is y² = 4ax.

DIFFERENT FORMS OF THE PARABOLA

Latus Rectum of the Parabola:

Let the given parabola be y² = 4ax.

In the figure LSL¢ is the latus rectum.

Also by definition, LSL¢ =2$\left(\sqrt{4a.a}\right)=4a$= double ordinate through the focus S.

Notes:

$\bullet$ Any chord of the parabola y² = 4ax which is perpendicular to its axis is called the double ordinate.

$\bullet$ Two parabolas are said to be equal when their latus recta are equal.

Illustration 1: Find the equation of parabola whose focus is (1, - 1), and whose vertex is (2, 1). Also find equation of its axis and latus rectum.

Solution: As we know that vertex is mid point of focus and point of intersection of directrix with the axis so point P(h, k) is $\frac{h+1}{2}=2\Rightarrow h=3$

$\frac{k-1}{2}=1\Rightarrow k=3$

$\therefore$ P is (3, 3)

slope of axis = $\frac{1-(-1)}{2-1}=2$

$\Rightarrow$ equation of axis = y – 1 = 2 (x – 2) $\Rightarrow$ 2x – y = 3

$\therefore$ equation of directrix y – 3 = -$\frac{1}{2}$(x – 3) $\Rightarrow$ x + 2y = 9

$\therefore$ equation of parabola is (x – 1)² + (y + 1)² = $\frac{{(x+2y-9)}^2}{5}$

equation of latus roctum

y + 1 = $\frac{1}{2}$(x – 1) $\Rightarrow$ x + 2y + 1 = 0

Parametric Equations of a Parabola:

If the coordinates of any point (x, y) on a curve can be expressed as functions of a variable t, given by x = $\varphi$ (t), y = $\psi$ (t) ……(1)

Then the equations in (1) are said to be parametric equations of the curve where 't' is called the parameter.

Clearly x = at², y = 2at satisfy the equation y² = 4ax for all real values of t. Hence the parametric equations of the parabola y² = 4ax are x = at², y = 2at, where t is the parameter.

Also, (at², 2at) is a point on the parabola y² = 4ax for all real values of t. This point is also described as the point 't' on the parabola.

Example -2: The tangents at the points (a$t_1^2$, 2at₁), (a$t_2^2$, 2at₂) on the parabola y² = 4ax are at right angles if

(A) t₁t₂ = -1 (B) t₁t₂ = 1

(C) t₁t₂ = 2 (D) t₁t₂ = -2

Solution: $\frac{1}{t_1}.\frac{1}{t_2}=-1$ $\Rightarrow$ t₁t₂ = -1

Hence (A) is the correct answer.

Any chord to the parabola y² = 4ax which passes through the focus is called a focal chord of the parabola y² = 4ax.

Let y² = 4ax be the equation of a parabola and (at², 2at) a point P on it. Suppose the coordinates of the other extremity Q of the focal chord through P are (at₁², 2at₁).

Then, PS and SQ, where S is the focus (a, 0) have the same slopes.

$\Rightarrow$ $\frac{2at-0}{a{t}^2-a}=\frac{2a{t}_1-0}{a{t_1}^2-a}$ $\Rightarrow$ tt₁² – t = t₁ t² - t₁ $\Rightarrow$ (tt₁ + 1)(t₁ – t) = 0.

Hence t₁ = -1/t, i.e. the point Q is (a/t², -2a/t).

i.e. the extremities of a focal chord of the parabola y² = 4ax may be taken as the points t and -1/t.

Example -3: Prove that the circle with any focal chord of the parabola y² = 4ax as its diameter always touches its directrix.

Solution: Let AB be a focal chord. If A is (at², 2at), then B is $\left(\frac{a}{t^2},-\frac{2a}{t}\right)$.

Equation of the circle with AB as diameter is

(x - at²)$\left(x-\frac{a}{t^2}\right)$ + (y - 2at)$\left(y+\frac{2a}{t}\right)$ = 0

For x = -a, this gives

$\frac{a^2{\left(1+t^2\right)}^2}{t^2}$ + y² - 2ay$\left(t-\frac{1}{t}\right)$ - 4a² = 0

$\Rightarrow$ a² ${\left(t-\frac{1}{t}\right)}^2$ + y² - 2ay(t - 1/t) = 0

$\Rightarrow$ [y - a(t - 1/t)]² = 0, which has equal roots.

$\Rightarrow$ x + a = 0 is a tangent to the circle with diameter AB.

Focal Distance of any Point:

The focal distance of any point P (x, y) on the parabola y² = 4ax is the distance between the point P and the focus S, i.e. PS.

Thus the focal distance

= PS = PM =ZN = ZA + AN = a + x

Position of a point relative to the Parabola:

Consider the parabola : y² = 4ax. If (x₁ , y₁) is a given point and y²₁ -4ax₁ = 0, then the point lies on the parabola. But when y₁² - 4ax₁ $\ne$ 0, we draw the ordinate PM meeting the curve in L. Then P will lie outside the parabola if

PM > LM, i.e., PM² – LM² > 0

Now, PM² = y₁² and LM² = 4ax₁ by virtue of the coordinates of L satisfying the equation of the parabola. Substituting these values in equation of parabola, the condition for P to lie outside the parabola becomes y₁² - 4ax₁ > 0.

Similarly, the condition for P to lie inside the parabola is y12 -4ax1 < 0.

Example -4:The coordinates of a point on the parabola y² = 8x, whose focal distance is 4, are

(A) $\left(\frac{1}{2},\pm 2\right)$ (B) (1, $\pm$2$\sqrt{2}$)

(C) (2, $\pm$4) (D) none of these

Solution: Focal distance of a point P (x, y) on y² = 4ax is (x + a).

$\therefore$ 4 = x + 2 $\Rightarrow$ x = 2

y² = 8 x 2 = 16 $\Rightarrow$ y = $\pm$ 4

Hence (C) is the correct answer.

General Equation of a Parabola

Let (h, k) be the focus S and lx + my + n = 0 the equation of the directrix ZM of a parabola. Let (x,y) be the coordinates of any point P on the parabola. Then the relation, PS = distance of P from ZM, gives

(x – h)² + (y – k)² = (lx + my + n)² / (l² + m²)

$\Rightarrow {(mx-ly)}^2+2gx+2fy+d=0$

This is the general equation of a parabola.

Note: From the general equation of the parabola it is clear that the second-degree terms in the equation of a parabola form a perfect square. The converse is also true, i.e. if in an equation of the second degree, the second-degree terms form a perfect square then the equation represents a parabola, unless it represents two parallel straight lines.

Special case:

Let the vertex be ($\alpha$, $\beta$) and the axis be parallel to the x-axis. Then the equation of parabola is given by (y - $\beta$)² = 4a (x – $\alpha$) which is equivalent to x = Ay² + By + C

Similarly, when the axis is parallel to the y-axis and vertex be($\alpha$, $\beta$), the equation of parabola is given by (x - $\alpha$)² = 4a (y – $\beta$) which is equivalent to y = A'x² + B'x + C'

Illustration 5: Find the vertex, axis, directrix, tangent at the vertex and the length of the latus rectum of the parabola 2y² + 3y - 4x - 3 = 0.

Solution: The given equation can be re-written as ${\left(y+\frac{3}{4}\right)}^2=2\left(x+\frac{33}{32}\right)$

which is of the form Y² = 4aX.

Hence the vertex is $\left(-\frac{33}{32},-\frac{3}{4}\right)$

The axis is $y+\frac{3}{4}=0$ $\Rightarrow$ y = -3/4

The directrix is x + a = 0

$\Rightarrow$ x + $\frac{33}{32}+\frac{1}{2}=0$$\Rightarrow$ x = -$\frac{49}{32}$

The tangent at the vertex is $\text{x}+\frac{33}{32}=0$ $\Rightarrow$ x = -$\frac{33}{32}$

Length of the latus rectum = 4a = 2.

Pole and Polar

To find the equation of the polar of the point (x₁, y₁) with respect to the parabola y² = 4ax.

Let Q and R be the points in which any chord drawn through the point P, whose coordinates are (x₁, y₁), meets the parabola.

Let the tangents at Q and R meet in the point whose coordiantes are (h, k).

We require the locus of (h, k).

Since QR is the chord of contact of tangents from (h, k) its equation is ky = 2a (x + h)

Since this straight line passes through the point (x₁, y₁) we have

ky₁ = 2a (x ₁+ h) ….(i)

Since the relation (i) is true, it follows that the point (h, k) always lies on the straight line

yy₁ = 2a(x + x₁) ….(ii)

Hence (ii) is the equation to the polar of (x₁, y₁).

Example -6:Prove that the locus of poles of focal chord of the parabola y² = 4ax is the directrix.

Solution: Let (h, k) be the pole. Then the equation of the chord is ky = 2a(x + h).

Since it is a focal chord, it passes through the focus (a, 0).

$\therefore$ 2a(a + h) = 0 or a + h = 0

Hence locus of pole (h, k) is

x + a = 0, which is the directrix.