Question

A tangent is drawn to the curve, $\frac{x^2}{16}+\frac{y^2}{9}=1$ at the point $P$ meeting the co-ordinate axis in $T$ and $t$ If OY is the perpendicular from the origin on the tangent then find the value of the product ( $Tt$ ) (PY).

Answer 1

Answer: 7

Tangent at $P:\frac{x\cos \theta }{a}+\frac{y\sin \theta }{b}=1$
T: $(a\sec \theta ,0),t:(0,b\mathrm{cosec}\theta )$
hence $Tt=\sqrt{a^2{\sec }^2\theta +b^2{\mathrm{cosec}}^2\theta }$
To find $Py$, we draw normal at $P$.
now $ON=Py$
Hence Normal at $P\frac{ax}{\cos \theta }-\frac{by}{\sin \theta }=a^2-b^2$
$ON:\frac{a^2-b^2}{\sqrt{a^2{\sec }^2\theta +b^2{\mathrm{cosec}}^2\theta }}=Py$
Hence Py. $Tt=a^2-b^2$