Q79 / 160 · AP EAMCET 2017 (26 Apr Shift 1) · Mathematics

If $l x+m y=1$ is a normal to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, then $a^2 m^2-b^2 l^2=$

  • A
    $\frac{m^2}{l^2}\left(a^2+b^2\right)^2$
  • B
    $\left(l^2+m^2\right)\left(a^2+b^2\right)^2$
  • C
    $\frac{l^2}{m^2}\left(a^2+b^2\right)^2$
  • Correct answer
    D
    $l^2 m^2\left(a^2+b^2\right)^2$

Solution

Given equation of normal, $l x+m y=1$ $$ y=\frac{-l}{m}+\frac{1}{m} $$ Which is of the form $y=m x+c$ Since, the lines $y=m x+c$ will be normal to the hyperbola $$ \begin{aligned} & \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \text { if } c^2=\frac{m^2\left(a^2+b^2\right)^2}{a^2-b^2 m^2} \\ & \therefore \quad\left(\frac{1}{m}\right)^2=\frac{\left(-\frac{l}{m}\right)^2\left(a^2+b^2\right)^2}{a^2-b^2\left(-\frac{l}{m}\right)^2} \end{aligned} $$ $\begin{aligned} & \frac{1}{m^2}=\frac{\frac{l^2}{m^2}\left(a^2+b^2\right)^2}{\frac{m^2 a^2-l^2 b^2}{m^2}} \\ & \frac{1}{m^2}=\frac{l^2\left(a^2+b^2\right)^2}{m^2 a^2-l^2 b^2} \\ & a^2 m^2-b^2 l^2=l^2 m^2\left(a^2+b^2\right)^2\end{aligned}$

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

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