Question

$P$ is a point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, $N$ is the foot of the perpendicular from $P$ on the transverse axis. The tangent to the hyperbola at $P$ meets the transverse axis at $T$. If $O$ is the centre of the hyperbola, then $OT$. $ON$ is equal to:

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

Answer 1

Answer: (B)

$a^2$

$P$ is a point on hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Let $P=(a\sec \theta ,b\tan \theta )$

$N$ is the foot of perpendicular on transverse axis

$\therefore N=(a\sec \theta ,0)$

Slope of tangent at $(a\sec \theta ,a\tan \theta )$ is $\frac{b}{a\sin \theta }$

Equation of tangent at point P is;

$y-b\tan \theta =\frac{b}{ad\sin \theta }(x-ad\sec \theta )$

Tangent meet transverse axis at $y=0$

Solving for $x$, we get ;

$\Rightarrow x=\frac{a}{\cos \theta }$

$\therefore T=(\frac{a}{\cos \theta },0)$

Here O=(0, 0)

$\therefore OT.ON=\left(a\cos \theta \right)\left(\frac{a}{d\cos \theta }\right)$

$\Rightarrow OT.ON=a^2$