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

The Correct Option is D

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 \((-3, -2)\).