Find the equation of parabola whose
Focus is $(3, 0)$ and the directrix is $3 x + 4 y = 1$

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Solution

focus $(3, 0)$

directrix $3 x + 4 y = 1$

Let A(x,y) be the pt. on the parabola

& AM be the length of $⊥$ from A to directrix

Then SA = AM

$\Rightarrow (S A)^{2} = (A M)^{2}$

$\Rightarrow (x - 3)^{2} + (y - 0^{2}) = {(\frac{(3 x + 4 y - 1)}{\sqrt{3^{2} + 4^{2}}})}^{2}$

$\Rightarrow (x - 3)^{2} + y^{2} = \frac{(3 x + 4 y - 1)^{2}}{25}$

$\Rightarrow x^{2} + 9 - 6 x + y^{2} = \frac{9 x^{2} + 16 y^{2} + 1 + 24 x y - 8 y - 6 x}{25}$

$\Rightarrow 25 x^{2} + 225 - 150 x + 25 y^{2} = 9 x^{2} + 16 y^{2} + 1 + 24 x y - 8 y - 6 x$

$\Rightarrow 16 x^{2} + 9 y^{2} - 24 x y - 14 y x + 8 y + 224 = 0$

Which is the required equation of parabola.

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