Question

Consider the ellipse $\frac{x^2}{3}+\frac{y^2}{1}=1$. Let P, Q, R, S be 4 points on this ellipse such that the normals drawn from these points are concurrent at (2, 2), then the centre of the conic (apart from the given ellipse) on which these 4 points lie is $(\alpha ,\beta )$. Find the value of $\frac{58}{(10\alpha +\beta )}$ is

Answer 1

Equation of normal at $(\sqrt{3}cos\theta ,sin\theta )$ will be
$\sqrt{3}xsin\theta -ycos\theta -2sin\theta cos\theta =0$
(if $h=\sqrt{3}cos\theta ;k=sin\theta )$
Line passing through (2, 2)
so , $\sqrt{3}sin\theta -cos\theta -sin\theta .cos\theta =0$
so, $\sqrt{3}k-\frac{h}{\sqrt{3}}-k.\frac{h}{\sqrt{3}}=0$
$\Rightarrow$ xy + x - 3y = 0 is a hyperbola
$Forcenter\equiv \frac{d}{dx}F(x,y)=0\Rightarrow y=-1$
$\frac{d}{dy}F(x,y)=0\Rightarrow x=3$
So, $(3,-1)$ is the centre

$(10\alpha +\beta )=29$

$\therefore \frac{58}{(10\alpha +\beta )}=2$