# What will be the equation of normal to the hyperbola 3x^2 – 4y^2 = 12 at the point (x1, y1)?

Category: QuestionsWhat will be the equation of normal to the hyperbola 3x^2 – 4y^2 = 12 at the point (x1, y1)?
Editor">Editor Staff asked 11 months ago

What will be the equation of normal to the hyperbola 3x^2 – 4y^2 = 12 at the point (x1, y1)?

(a) 3x1y + 4y1x + 7x1y1 = 0

(b) 3x1y + 4y1x – 7x1y1 = 0

(c) 3x1y – 4y1x – 7x1y1 = 0

(d) 3x1y – 4y1x + 7x1y1 = 0

This question was posed to me by my school principal while I was bunking the class.

My question is based upon Calculus Application topic in section Application of Calculus of Mathematics – Class 12
NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options
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Editor">Editor Staff answered 11 months ago

Correct answer is (b) 3x1y + 4y1x – 7x1y1 = 0

The explanation: Equation of the given hyperbola is, 3x^2 – 4y^2 = 12 ……….(1)

Differentiating both sides of (1) with respect to y we get,

3*2x(dy/dx) – 4*(2y) = 0

Or dx/dy = 4y/3x

Therefore, the equation of the normal to the hyperbola (1) at the point (x1, y1) on it is,

y – y1 = -[dx/dy](x1, y1) (x – x1) = -4y1/3×1(x – x1)

Or 3x1y + 4y1x – 7x1y1 = 0