If the equation of the parabola, whose vertex is at (5, 4) and the...

Given the equation of the parabola whose vertex is $\left(5, 4\right) and equation of directrix is$3x + y - 29 = 0$is${x}^{2} + a{y}^{2} + bxy + cx + dy + e = 0$</p><p>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$</p><p>Now the intersection of$3x + y - 29 = 0 \text{ and } x - 3y + 7 = 0$will be$Z = \left(8, 5\right)$</p><p>Now plotting the diagram we get,</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/93cb329f-f8e1-4906-85ee-f88cbac96d43-image.png" style="width: 250px; height: 175px;" /></p><p>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$⇒ \dfrac{\alpha + 8}{2} = 5, \dfrac{\beta + 5}{2} = 4 ⇒ \left(\alpha, \beta\right) = \left(2, 3\right)$

$∴$ Focus is \left(2, 3\right)$and directrix is$3x + y - 29 = 0$</p><p>Applying${\left(PS\right)}^{2} = {\left(PM\right)}^{2}$, we get</p><p>${\left(x - 2\right)}^{2} + {\left(y - 3\right)}^{2} = \dfrac{{\left(3x + y - 29\right)}^{2}}{10}$</p><p>$⇒ {x}^{2} + 9{y}^{2} - 6xy + 134x - 2y - 711 = 0$</p><p>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$