Question

A normal is drawn at a point $(x_1,y_1)$ of the parabola $y^2=16x$ and this normal makes equal angle with both $X$ and $Y$-axes. Then, point $(x_1,y_1)$ is

• A. $(4,-8)$
• B. $(1,-4)$
• C. $(4,-4)$
• D. $(2,-8)$

Answer 1

The correct option is A $(4,-8)$

Given equation of parabola is
$y^2=16x$
On differentiating both sides, we get
$2y{y}^{{}^{\prime }}=16$
$\Rightarrow y^{{}^{\prime }}=\frac{16}{2y}=\frac{8}{y}$
$\therefore$ Slope of tangent at point $(x_1,y_1),m_1=\frac{8}{y_1}$
and slope of normal at point $(x_1,y_1),m_2=\frac{-y_1}{8}$
Since, normal makes equal angle with both $X$ and $Y$-axes, then
$m_2=\pm 1$
$\Rightarrow \frac{-y_1}{8}=\pm 1$
$\Rightarrow -y_1=\pm 8$
Now, when $y_1=8$, then $x_1=4$
when $y_1=-8$, then $x_1=4$
So, the required point is $(4,-8)$.