Question

The line $y=mx+\sqrt{a^2{m}^2-b^2}$ touches the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ at the point $P(a\sec \theta ,b\tan \theta )$ then $\theta$ is

• A. ${\sin }^{-1}\left(\frac{b}{am}\right)$
• B. ${\sin }^{-1}\left(\frac{a}{bm}\right)$
• C. ${\cos }^{-1}\left(\frac{b}{am}\right)$
• D. $Noneofthese$

Answer 1

Answer: (C)

${\cos }^{-1}\left(\frac{b}{am}\right)$

The slope of the tangent at any point on the hyperbola can be calculated by differentiating it w.r.t. $x.$

$\frac{2x}{a^2}-\frac{2y\frac{dy}{dx}}{b^2}=0$

$\Rightarrow \frac{dy}{dx}=\frac{b^{2}x}{a^{2}y}$

$\Rightarrow m=\frac{b^{2}a\sec \theta }{a^{2}b\tan \theta }=\frac{b}{a\sin \theta }$

$\Rightarrow \sin \theta =\frac{b}{am}$