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

Answer 1

Answer: (C)

$(4, - 8)$

Answer 2

Given equation of parabola is
$y^{2}=16 x$
On differentiating both sides, we get
$2 y y'=16$
$y'=\frac{16}{2 y}=\frac{8}{y}$
$\therefore$ Slope of tangent at point $\left(x_{1}, y_{1}\right), m_{1}=\frac{8}{y_{1}}$
and slope of normal at point $\left(x_{1}, y_{1}\right), m_{2}=\frac{-y_{1}}{8}$
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 $(4,-8)$