Q95 / 160 · AP EAMCET 2022 (05 Jul Shift 2) · Mathematics
The locus of the point of intersection of the tangents at the
endpoints of normal chords of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is
- A$\frac{a^6}{x^2}+\frac{b^6}{y^2}=\left(a^2+b^2\right)^2$
- Correct answerB$\frac{a^6}{x^2}-\frac{b^6}{y^2}=\left(a^2+b^2\right)^2$
- C$\frac{a^6}{x^2}-\frac{b^6}{y^2}=\left(a^2-b^2\right)^2$
- D$\frac{a^6}{x^2}+\frac{b^6}{y^2}=\left(a^2-b^2\right)^2$
Solution
Let $p(h, k)$ be point of intersection of tangents at end points of normal chord
Standard equation of normal chord of hyperbola
$a x \cos \theta+b y \cot \theta=a^2+b^2$ ...(i)
for $p(h, k)$, equation of chord of contact of tangents to hyperbola is
$\frac{h x}{a^2}-\frac{k y}{b^2}=1$ ...(ii)
(i) and (ii) represent some line
$\frac{a \cos \theta}{\left(\frac{h}{a^2}\right)}=\frac{b \cot \theta}{\left(\frac{-k}{b^2}\right)}=\frac{a^2+b^2}{1}$
$\frac{a^3 \cos \theta}{h}=\frac{-b^3 \cot \theta}{k}=\frac{a^2+b^2}{1}$
$\begin{aligned} \cos \theta & =\frac{h\left(a^2+b^2\right)}{a^3} \\ \sec \theta & =\frac{a^3}{h\left(a^2+b^2\right)}\end{aligned}$
$\begin{aligned} & \text { and } \cot \theta=\frac{-k\left(a^2+b^2\right)}{b^3} \\ & \tan \theta=\frac{-b^3}{k\left(a^2+b^2\right)}\end{aligned}$
$\sec ^2 \theta-\tan ^2 \theta=1$
$\begin{aligned} & \frac{a^6}{h^2\left(a^2+b^2\right)^2}-\frac{b^6}{k^2\left(a^2+b^2\right)^2}=1 \\ & \frac{a^6}{h^2}-\frac{b^6}{k^2}=\left(a^2+b^2\right)^2 \\ & \frac{a^6}{x^2}-\frac{b^6}{y^2}=\left(a^2+b^2\right)^2\end{aligned}$
Hence, the correct answer is option: 'B'.