Question

A normal to the curve $2x^{2 }- y^{2} = 14$ at the point $(x_{1}, y_{1})$ is parallel to the straight line $x + 3y = 4$ . Then the point $(x_{1}, y_{1})$ is

• A. $(3,3)$
• B. $(-4,-2)$
• C. $(2,3)$
• D. $(3,2)$

Answer 1

Answer: (D)

$(3,2)$

Answer 2

The correct answer is D:(3,2)
Given curve is $2x^2 - y^2 = 14 \,\,\,... (i)$
Differentiating $(i)$ w.r.t. $x$, we get
$4x - 2y \frac{dy}{dx} = 0$
$\Rightarrow \frac{dy}{dx} = \frac{2x}{y}$
Let $m_1 =$ Slope of normal at $(x_1, y_1) = \frac{-y_1}{2x_1}$
Also, slope of line $x + 3y = 4$ is $m_2=\frac{-1}{3}$
Since the line and normal to the curve are parallel.
$\therefore m_1 = m_2$
$\Rightarrow \frac{-y_1}{2x_1} = \frac{-1}{3}$
$\Rightarrow 2x_1 = 3y_1$
$\Rightarrow y_1 = \frac{2}{3} x_1$
Since $(x_1, y_1)$ lies on $(i)$
$\therefore 2x_1^2 - y_1^2 = 14$
$\Rightarrow 2x_1^2 - \frac{4}{9} x_1^2 = 14$
$\Rightarrow 14x^2_1 = 126$
$\Rightarrow x_1 = \pm 3$
$\therefore y_1 = 2$ or $-2$
So, point $(x_1, y_1)$ is $(3, 2)$ or $(-3, -2)$.