If the equation of the parabola, whose vertex is at 5 , 4 and the directrix is 3 x + y - 29 = 0 , is x 2 + a y 2 + b x y + c x + d y + k = 0 , then a + b + c + d + k is equal to

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Aicharya · Accepted Answer

Given the equation of the parabola whose vertex is $(5, 4)$ and equation of directrix is $3 x + y - 29 = 0$ is $x^{2} + a y^{2} + b x y + c x + d y + e = 0$

Let focus be $(α, β)$

Let equation of $Z V$ which is perpendicular to $3 x + y - 29 = 0$ be $x - 3 y + k = 0$ and it passes through point $(5, 4)$ . So, $k = 7$ and equation of $Z V$ become $x - 3 y + 7 = 0$

Now the intersection of $3 x + y - 29 = 0 & x - 3 y + 7 = 0$ will be $Z = (8, 5)$

Now plotting the diagram we get,

So, foot of perpendicular from $(5, 4)$ on $3 x + y - 29 = 0$ is $(8, 5)$

Now using the midpoint formula we will find the focus of parabola $\Rightarrow \frac{α + 8}{2} = 5, \frac{β + 5}{2} = 4 \Rightarrow (α, β) = (2, 3)$

$∴$ Focus is $(2, 3)$ and directrix is $3 x + y - 29 = 0$

Applying ${(P S)}^{2} = {(P M)}^{2}$ , we get

${(x - 2)}^{2} + {(y - 3)}^{2} = \frac{{(3 x + y - 29)}^{2}}{10}$

$\Rightarrow x^{2} + 9 y^{2} - 6 x y + 134 x - 2 y - 711 = 0$

Comparing with $x^{2} + a y^{2} + b x y + c x + d y + e = 0$ , we get $(a + b + c + d + e) = (9 - 6 + 134 - 2 - 711) = - 576$

If the equation of the parabola, whose vertex is at 5 , 4 and the directrix is 3 x + y - 29 = 0 , is x 2 + a…

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