Question

Foci are $(3,2)$ and $(1,-2)$ and major axis is of length $10.$ If the equation of ellipse is $p{x}^2+qxy+21y^2-96x+ry-404=0$. Find the value of $p+q+r.$

• A. $25$
• B. $27$
• C. $28$
• D. $30$

Answer 1

The correct option is C $28$

We know that in the case of an ellipse the sum of the focal distances of any point on the ellipse is equal to length of major axis. $\therefore \sqrt{\{{(x-3)}^2+{(y-2)}^2\}}+\sqrt{\{{(x-1)}^2+{(y+2)}^2\}}=\mathrm{10...}\left(1\right)$.

On simplification we find that
$\{{(x-3)}^2+{(y-2)}^2\}-\{{(x-1)}^2+{(y+2)}^2\}=-4x-8y+\mathrm{8...}\left(2\right)$
Dividing (2) by (1) and noting that
$\frac{a^2-b^2}{a+b}=a-b$,
$\sqrt{\left\{\right({x-3)}^2+\left({y-2)}^2\right\}}-\sqrt{\left\{\right({x-1)}^2+\left({y+2)}^2\right\}}=-\frac{4(x+2y-2)}{10}$..$\left(3\right)$
Adding (1) and (3), we get
$2\sqrt{\{{(x-3)}^2+{(y-2)}^2\}}=10-\frac{2}{5}.(x+2y-2)$
$25(x^2+y^2-6x-4y+13)={(27-x-2y)}^2$ or
$24x^2-4xy+21y^2-96x+8y-404=0$.
Then,$p+q+r=24-4+8=28$.