Question

The locus of the point of intersection of the tangents at the points with eccentric angles $\varphi$ and $\frac{\pi }{2}-\varphi$ on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is

• A. $x=a$
• B. $y=b$
• C. $x=ab$
• D. $y=ab$

Answer 1

Answer: (B)

$y=b$

Let $P(a\sec \varphi ,b\tan \varphi )$ and $Q(a\csc \varphi ,b\cot \varphi )$ be two points with eccentric angles $\varphi$ and $\frac{\pi }{2}-\varphi$ on $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
The equations of tangents at $P$ and $Q$ are $\frac{x\sec \varphi }{a}-\frac{y\tan \varphi }{b}=1$ and $\frac{x\csc \varphi }{a}-\frac{y\cot \varphi }{b}=1$ respectively.
Eliminating $\varphi$ from the above equations
$\csc \varphi \left(\frac{x\sec \varphi }{a}-\frac{y\tan \varphi }{b}=1\right)$ --------(1)
$\sec \varphi \left(\frac{x\csc \varphi }{a}-\frac{y\cot \varphi }{b}=1\right)$ ---------(2)
(1) - (2) gives $y=b$
Hence, the required locus is $y=b$.