Question

lf the chord joining the points $P\left(\alpha \right)$, $Q\left(\beta \right)$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ subtends a right angle at its centre then $\tan \alpha \tan \beta =$

• A. $\frac{-a^2}{b^2}$
• B. $\frac{a^2}{b^2}$
• C. $\frac{-b^2}{a^2}$
• D. $\frac{b^2}{a^2}$

Answer 1

The correct option is A $\frac{-a^2}{b^2}$

To find value of $tan\alpha tan\beta$
Parametric form of Points are
$P(acos\alpha ,bsin\alpha )$, $Q(-asin\beta ,bcos\beta )$
Since OP and OQ are Perpendicular
OP.OQ = 0
$acos\alpha [-asin\beta ]+bsin\alpha \left[bcos\beta \right]=0$
On Simplifying this , we get
$(b^2+a^2)Cos(\alpha -\beta )=(b^2-a^2)Cos(\alpha +\beta )$
$\frac{b^2+a^2}{b^2-a^2}=\frac{Cos(\alpha +\beta )}{Cos(\alpha -\beta )}$
$\frac{b^2}{a^2}=\frac{Cos(\alpha +\beta )+Cos(\alpha -\beta )}{Cos(\alpha +\beta )-Cos(\alpha -\beta )}$
Finally , We get
$tan\alpha tan\beta =\frac{-a^2}{b^2}$
Hence , Option A