Question

The normal to curve $xy=4$ at the point $(1,4)$ meets the curve again at point

• A. $(-4,-1)$
• B. $(-8,-\frac{1}{2})$
• C. $(-16,-\frac{1}{4})$
• D. $(-1,-4)$

Answer 1

The correct option is C $(-16,-\frac{1}{4})$

Let the point $P$ be $(2t_1,\frac{2}{t_1})$
Slope of the normal through $P$ is $t_1^2\cdots \left(1\right)$
Let the normal at $P$ meet again at $Q(2t_2,\frac{2}{t_2})$
Slope of the normal at $QP=\frac{2/t_2-2/t_1}{2t_2-2t_1}=\frac{-1}{t_1{t}_2}\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right)$,
$t_1^3.t_2=-1$

Given point is $P(1,4)$
$\Rightarrow t_1=1/2;t_2=-8$
Thus $Q$ will be $(-16,-\frac{1}{4})$