Question

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

• A. $3$
• B. $9$
• C. $-9$
• D. $-3$

Answer 1

Answer: (B)

Let $x+y=k$ be normal to
$y^2=12x$ at $P(\alpha ,\beta )$...(i)
$\therefore {\beta }^2=12\alpha$...(ii)
Also, slope of normal at $P(\alpha ,\beta )$ is -1
Form, Eq.(i),
$\frac{dy}{dx}=\frac{6}{y}$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_P=\frac{6}{\beta }$
$\therefore -1=\frac{-1}{6/\beta }$
$\Rightarrow \beta =6,\alpha =3$
$\therefore P$ is (3,6) which lies on $x+y=k$
$\therefore 3+6=k$
$\Rightarrow k=9$