Question

Find the equation of parabola having vertex (2,-3) and directrix x-4y-48=0

Answer 1

Given vertex A(2,-3) and directrix is x - 4y - 48 = 0.

$Slope \:of \: the \:directrix = \frac{1}{4}$

$Slope \:of \: the \:axis= -4$

The equation of axis passing through point A and having slope -4 is:

y + 3 = -4(x-2)

=> y + 3 = -4x + 8

=> 4x + y -5 = 0

$Z \:is \: the \:foot \:of \:the \: perpendicular \\from \: A(2,-3) \:to \:the \: directrix \:x-4y-48 = 0$

$\frac{h-2}{1} = \frac{k+3}{-4} = \frac{-(2+12-48}{1+16}$

$\implies \frac{h-2}{1} = \frac{k+3}{-4} = \frac{34}{17}$

$\implies \frac{h-2}{1} = \frac{k+3}{-4} = 2$

$i) h - 2 = 2 \implies h = 4$

$ii ) \frac{k+3}{-4} = 2 \\ \implies k + 3 = -8 \\\implies k = -11$

$A \:is \: the \: mid-point \: of \: SZ$

$(2,-3) = \Big( \frac{4+p}{2} , \frac{-11+q}{2}\Big)$

$iv) 2 = \frac{4+p}{2} \\\implies 4 = 4 + p \\\implies p = 0$

$v) -3 = \frac{-11+q}{2} \\\implies -6 = -11 + q \\\implies q = 5$

$\therefore S = (0,5)$

By definition of Parabola :

SP = PM

$(x-0)^{2}+(y-5)^{2} = \Big(\frac{x-4y-48}{\sqrt{1+16}}\Big)^{2}$

$\implies x^{2}+y^{2}+25-10y \\= \frac{x^{2}+16y^{2}+2304-8xy+384y-96x}{17}$

$\implies 17x^{2}+17y^{2}+426-170y \\= x^{2}+16y^{2}+2304-8xy+384y-96x$

Therefore.,

$\red { Required \: equation}$

$\green { 16x^{2}+8xy+y^{2}+96x-208y-1879 =0}$

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