# What is the nature of the straight line x + y + 7 = 0 to the hyperbola 3x^2 – 4y^2 = 12 whose normal is at the point (x1, y1)?

Category: QuestionsWhat is the nature of the straight line x + y + 7 = 0 to the hyperbola 3x^2 – 4y^2 = 12 whose normal is at the point (x1, y1)?
Editor">Editor Staff asked 11 months ago

What is the nature of the straight line x + y + 7 = 0 to the hyperbola 3x^2 – 4y^2 = 12 whose normal is at the point (x1, y1)?

(a) Chord to hyperbola

(b) Tangent to hyperbola

(c) Normal to hyperbola

(d) Segment to hyperbola

My enquiry is from Calculus Application in division Application of Calculus of Mathematics – Class 12
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Editor">Editor Staff answered 11 months ago

The correct choice is (c) Normal to hyperbola

The best 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

Now, if possible, let us assume that the straight line

x + y + 7 = 0 ………..(2)

This line is normal to the hyperbola (1) at the point (x1, y1). Then, the equation (2) and (3) must be identical. Hence, we have,

3×1/1 = 4y1/1 = -7x1y1/7

So, x1 = -4 and y1 = -3

Now, 3×1^2 – 4y1^2 = 3(-4)^2 – 4(-3)^2 = 12

This shows the point (-4, -3) lies on the hyperbola (1).

Thus, it is evident that the straight line (3) is normal to the hyperbola (1).