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

I had been asked this question in final exam.

My enquiry is from Calculus Application in division Application of Calculus of Mathematics – Class 12

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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).