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
NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options
Interview Questions and Answers, Database Interview Questions and Answers for Freshers and Experience
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).