Find the equation of parabola whose focus is the point $(2, 3)$ and directrix is the line $x - 4 y + 3 = 0$ . Also, find the length of its latus-rectum.

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Solution

Let A(x,y) be the Pt. on the parabola
focus = (2,3) & $e q^{n}$ of directrix $= x - 4 y + 3 = 0$
Now let AM be a perpendicular from A on the directrix.
then, we have $(S A)^{2} = (A M)^{2}$
$(x - 2)^{2} + (y - 3)^{2} = (\frac{x - 4 y + 3}{\sqrt{1^{2} + (- 4)^{2}}})^{2}$
$x^{2} - 4 x + 4 + y^{2} - 6 y + 9 = \frac{(x + 4 y + 3)^{2}}{17}$
$17 x^{2} + 17 y^{2} - 68 x - 102 y + 221 = x^{2} + 16 y^{2} + 9 - 8 x y + 6 x - 24 y$
$\Rightarrow 16 x^{2} + y^{2} + 8 x y - 74 x - 78 y + 212 = 0$
Which is the required $e q^{n}$ of parabola.
Latus rectum = $2 \times$ (length of $⊥$ from focus in th directrix)
$= 2 \times ∣ ∣ ∣ ∣ \frac{(2 - 2 \times 3 + 3)}{\sqrt{1^{2} + 4^{2}}} ∣ ∣ ∣ ∣$
$\Rightarrow 2 \times ∣ ∣ ∣ \frac{- 7}{\sqrt{1 + 16}} ∣ ∣ ∣ \Rightarrow \frac{+ 14}{\sqrt{17}}$
Latus Rectum $= \frac{14}{\sqrt{17}}$

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Similar questions

(2, 3)

x - 4 y + 3 = 0.

(- 2, 3)

2 x + 3 y - 4 = 0

Length of latus rectum of the parabola whose focus is at $(2, 3)$ and directrix is the line $x - 4 y + 3 = 0$ is

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x = - 3

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