Name: Normals to the Parabolas
Uploaded: 2019-07-19
Duration: 9 min 41 s
Description: This video will give a clear understanding of how to write and learn the concept of normal to the parabola in the parametric form. Don't miss it.

Question

Find the equation of the parabola, if the focus is at $(- 6, - 6)$ and the vertex is at $(- 2, 2)$ .

Accepted Answer

In a parabola, the vertex is the mid-point of the focus and the point of intersection of the axis and the directrix.

Let $(x_{1}, y_{1})$ be the coordinates of the point of intersection of the axis and directrix.

It is given that the vertex and the focus of a parabola are $(- 2, 2)$ and $(- 6, - 6)$ , respectively.

$∴$ Slope of the axis of the parabola $= \frac{- 6 - 2}{- 6 + 2} = \frac{- 8}{- 4} = 2$

Slope of the directrix $= \frac{- 1}{2}$

Let the directrix intersect the axis at $K (r, s)$ .

$∴ \frac{r - 6}{2} = - 2, \frac{s - 6}{2} = 2$

$\Rightarrow r = 2, s = 10$

$∴$ Required equation of the directrix.

$y - 10 = \frac{- 1}{2} (x - 2)$

$\Rightarrow 2 y + x - 22 = 0$

Now, let $P (x, y)$ be any point on the parabola whose focus is $S (- 6, - 6)$ and the directrix is

$2 y + x - 22 = 0$ .

Draw $P M$ perpendicular to $2 x + y + 22 = 0$

Then, we have:

$SP = PM$

$\Rightarrow S P^{2} = P M^{2}$

$\Rightarrow (x + 6)^{2} + (y + 6)^{2} = {(\frac{2 y + x - 22}{\sqrt{5}})}^{2}$

$\Rightarrow 5 (x^{2} + 12 x + 36 + y^{2} + 12 y + 36) = 4 y^{2} + x^{2} + 484 + 4 x y - 88 y - 44 x$

$\Rightarrow 4 x^{2} + y^{2} - 4 x y + 104 x + 148 y - 124 = 0$

$\Rightarrow (2 x - y)^{2} - 4 (26 x + 37 y - 31) = 0$

4.2.1

K. C. Sinha Mathematics Solutions for Exercise - 4.2.1

Questions from 4.2.1 with Hints & Solutions

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