Q81 / 150 · MHT CET 2011 · Mathematics

If $x+y=k$ is normal to $y^{2}=12 x$, then $k$ is

  • A
    3
  • Correct answer
    B
    9
  • C
    $-9$
  • D
    $-3$

Solution

Let $x+y=k$ be normal to $$ y^{2}=12 x \text { at } P(\alpha, \beta) $$ $\therefore \quad \beta^{2}=12 \alpha$ Also, slope of normal at $P(\alpha, \beta)$ is $-1$ Form, Eq.(i), $$ \begin{aligned} \frac{d y}{d x} &=\frac{6}{y} \\ \Rightarrow \quad\left(\frac{d y}{d x}\right)_{P} &=\frac{6}{\beta} \end{aligned} $$ $\therefore$ $$ -1=\frac{-1}{6 / \beta} $$ $\Rightarrow$ $\beta=6, \alpha=3$ $\therefore P$ is $(3,6)$ which lies on $x+y=k$ $\therefore \quad 3+6=k$ $\Rightarrow \quad k=9$

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

Relevant Questions