Q98 / 160 · TS EAMCET 2016 · Mathematics
Let $A(2 \sec \theta, 3 \tan \theta)$ and $B(2 \sec \phi, 3 \tan \phi)$ where $\theta+\phi=\frac{\pi}{2}$, be two points on the hyperbola $\frac{x^2}{4}-\frac{y^2}{9}=1$. If $(\alpha, \beta)$ is the point of intersection of normals to the hyperbola at $A$ and $B$, then $\beta$ is equal to
- Correct answerA$-\frac{13}{3}$
- B$\frac{13}{3}$
- C$\frac{3}{13}$
- D$-\frac{3}{13}$
Solution
The normal at point $A(2 \sec \theta, 3 \tan \theta)$ is
$$
\begin{aligned}
2 x \cos \theta+3 y \cot \theta=4+9 \\
2 x \cos \theta+3 y \cot \theta=13
\end{aligned}
$$
And the normal at point $B(2 \sec \phi, 3 \tan \phi)$ is
$$
\begin{aligned}
& 2 x \cos \phi+3 y \cot \phi=4+9 \\
& \Rightarrow \quad 2 x \cos \phi+3 y \cot \phi=13 \\
&
\end{aligned}
$$
Multiplying Eq. (i) by $\cos \phi$ and Eq. (ii) by $\cos \theta$, then subtracting Eq. (ii) from Eq. (i), we get
$$
\begin{aligned}
& 3 y \cot \theta \cos \phi-3 y \cot \phi \cos \theta=13(\cos \phi-\cos \theta) \\
& \Rightarrow 3 y\left\{\cot \left(\frac{\pi}{2}-\phi\right) \cos \phi-\cot \phi \cos \left(\frac{\pi}{2}-\phi\right)\right\}
\end{aligned}
$$
$$
\begin{array}{rlrl}
& & =13\left\{\cos \phi-\cos \left(\frac{\pi}{2}-\phi\right)\right\} \\
\Rightarrow & 3 y\{\tan \phi \cos \phi-\cot \phi \sin \phi\}=13(\cos \phi-\sin \phi) \\
\Rightarrow & 3 y(\sin \phi-\cos \phi) & =13(\cos \phi-\sin \phi) \\
\Rightarrow & & 3 y & =\frac{-13(\sin \phi-\cos \phi)}{\sin \phi-\cos \phi} \\
& & y & =-\frac{13}{3}
\end{array}
$$
Hence, the value of $\beta$ is $-\frac{13}{3}$.
Hence, the correct answer is option: 'A'.