Q98 / 160 · AP EAMCET 2007 · Mathematics

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

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

Solution

If $l x+m y+n=0$ is normal to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ Then $\frac{a^2}{l^2}-\frac{b^2}{m^2}=\frac{\left(a^2+b^2\right)^2}{n^2}$ Here, $n=-1$, therefore $\frac{a^2}{l^2}-\frac{b^2}{m^2}=\left(a^2+b^2\right)^2$.

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

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