Q98 / 160 · TS EAMCET 2018 (04 May Shift 2) · Mathematics

If the product of the slopes of the tangents drawn from an external point $P$ to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is a constant $k^2$, then the locus of $P$ is

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

Solution

Let the point be $P(h, k)$. Tangent from $(h, k)$ to hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $\left(\frac{h x}{a^2}-\frac{k y}{b^2}-1\right)^2=\left(\frac{h^2}{a^2}-\frac{k^2}{b^2}-1\right)$ $$ \left(\frac{x^2}{a^2}-\frac{y^2}{b^2}-1\right) $$ $$ \begin{array}{cc} \Rightarrow & \frac{h^2 x^2}{a^4}+\frac{k^2 y^2}{b^4}+1-\frac{2 h k x y}{a^2 b^2}-\frac{2 h x}{a^2}+\frac{2 k y}{b^2} \\ = & \frac{h^2 x^2}{a^4}-\frac{h^2 y^2}{a^2 b^2}-\frac{h^2}{a^2}-\frac{x^2 k^2}{a^2 b^2}+\frac{y^2 k^2}{b^4}+\frac{k^2}{b^2}-\frac{x^2}{a^2}+\frac{y^2}{b^2}+1 \\ \Rightarrow \quad & \frac{x^2}{a^2}\left(\frac{k^2}{b^2}+1\right)-\frac{2 h k x y}{a^2 b^2}+\frac{y^2}{b^2} \\ & \left(\frac{h^2}{a^2}-1\right)-\frac{2 h x}{a^2}+\frac{2 k y}{b^2}+\frac{h^2}{a^2}=0 \\ \therefore & m_1 m_2=\frac{k^2+b^2}{h^2-a^2} \end{array} $$ Product of shape $=k^2$ $\therefore$ Locus of $p$ is $\frac{y^2+b^2}{x^2-a^2}=k^2$ $$ \Rightarrow \quad y^2+b^2=k^2\left(x^2-a^2\right) $$

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

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