Question

The equation of the normal at the point (2, √7) to the hyperbola 16x² - 9y² = 1 is:
1. \(32y + 9\sqrt 7 x = \sqrt 7 \)
2. \(32y + 9\sqrt 7 x = 50\sqrt 7 \)
3. \(32y + 9\sqrt 7 x = 1\)
4. \(32y + 9\sqrt 7 x = 2\sqrt 7\)

Answer 1

Correct Answer - Option 2 : \(32y + 9\sqrt 7 x = 50\sqrt 7 \)

Calculation:

Given equation of Hyperbola 16x2 - 9y2 = 1

or differentiating both side with respect to 'x'.

\(16 \times 2x - 9 \times 2y \dfrac{dy}{dx}=0\)

\(⇒ 32x - 18y \dfrac{dy}{dx} = 0 \)

\(⇒ 18y \dfrac{dy}{dx} = 32 x \)

\(⇒ \dfrac{dy}{dx} = \dfrac{32x}{18y} = \dfrac{16x}{9y} ⇒ \dfrac{dy}{dx} = \dfrac{16x}{9y}\)

\(m= \left(\dfrac{dy}{dx}\right)_{(2, √7)} = \left(\dfrac{16\times 2}{9\times √ 7}\right)=\dfrac{32}{9√ 7}\)

Equation of normal at (2, √7)

\(\Rightarrow y - √ 7 = - \dfrac{1}{m} (x - 2)\)

\(\Rightarrow y - √ 7 = - \dfrac{1}{\dfrac{32}{9√ 7}}(x - 2)\)

\(\Rightarrow y - √ 7 = -\dfrac {9√ 7}{32} ( x - 2) \)

⇒ 32y - 32√7 = -9√7x + 18√7

⇒ 9√7x + 32y = 50√7