JEE Main 2022 — Parabola Question with Solution

Given the equation of the parabola whose vertex is $\left(5,4\right)$ and equation of directrix is $3x+y-29=0$ is $x^2+a{y}^2+bxy+cx+dy+e=0$

Let focus be $\left(\alpha ,\beta \right)$

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

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

Now plotting the diagram we get,

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

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

$\therefore$ Focus is $\left(2,3\right)$ and directrix is $3x+y-29=0$

Applying ${\left(PS\right)}^2={\left(PM\right)}^2$, we get

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

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

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