Q94 / 160 · AP EAMCET 2022 (05 Jul Shift 1) · Mathematics

The locus of point of intersection of tangents at the ends of normal chord of the hyperbola $x^2-y^2=a^2$ is

  • A
    $y^4-x^4=4 a^2 x^2 y^2$
  • B
    $y^2-x^2=4 a^2 x^2 y^2$
  • Correct answer
    C
    $a^2\left(y^2-x^2\right)=4 x^2 y^2$
  • D
    $y^2+x^2=4 a^2 x^2 y^2$

Solution

Let $p(h, k)$ be the point of intersection of tangents at the ends of a normal chord of the hyperbola, $x^2-y^2=a^2$, then the equation of the chord is $h x-k y=a^2$ ...(i) But its is normal chord. So, its equation must be of the form $x \cos \theta+y \cot \theta=2 a$ ...(ii) Eqs. (i) and (ii) represents the same line $\therefore \quad \frac{\cos \theta}{h}=\frac{\cot \theta}{-k}=\frac{2 a}{a^2}$ $\sec \theta=\frac{a}{2 h}, \tan \theta=\frac{-a}{2 k}$ $\therefore \quad \sec ^2 \theta-\tan ^2 \theta=1$ $\begin{array}{ll}\Rightarrow & \frac{a^2}{4 h^2}-\frac{b^2}{4 k^2}=1 \\ \Rightarrow & a^2\left(k^2-h^2\right)=4 h^2 k^2\end{array}$ Therefore, the locus of $p(h, k)$ is $a^2\left(y^2-x^2\right)=4 x^2 y^2$

Hence, the correct answer is option: 'C'.

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