Question

The locus of the point of intersection of the tangents at the end points of normal chords of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$:

• A. $\frac{a^6}{x^2}+\frac{b^6}{y^2}={(a^2+b^2)}^2$
• B. $\frac{a^6}{x^2}-\frac{b^6}{y^2}={(a^2+b^2)}^2$
• C. $\frac{a^6}{x^2}-\frac{b^6}{y^2}={(a^2-b^2)}^2$
• D. $\frac{a^6}{x^2}+\frac{b^6}{y^2}={(a^2-b^2)}^2$

Answer 1

Answer: (B)

$\frac{a^6}{x^2}-\frac{b^6}{y^2}={(a^2+b^2)}^2$

Let $P(h,k)$ be the point of tangents at the end-points of a normal chord
$ax\cos \theta -by\cot \theta =a^2+b^2\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(i\right)$
The equation of chord of contact of tangents drawn from $P(h,k)$ to the hyperbola is
$\frac{{hx}^{}}{a^2}-\frac{{ky}^{}}{b^2}=1\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(ii\right)$
Clearly $\left(i\right)$ and $\left(ii\right)$ represent the same line
$\therefore \frac{a^3\cos \theta }{h}=-\frac{b^3\cot \theta }{k}=\frac{a^2+b^2}{1}$
$\Rightarrow \sec \theta =\frac{a^3}{h(a^2+b^2)},\tan \theta =-\frac{b^3}{k(a^2+b^2)}$
$\Rightarrow {\sec }^2\theta -{\tan }^2\theta =\frac{a^6}{h^2{(a^2+b^2)}^2}-\frac{b^6}{k^2{(a^2+b^2)}^2}$
$\Rightarrow \frac{a^6}{h^2}-\frac{b^6}{k^2}={(a^2+b^2)}^2$
Therefore, the locus of $P(h,k)$ is $\frac{a^6}{x^2}-\frac{b^6}{y^2}={(a^2+b^2)}^2$
Hence, option 'B' is correct.