JEE Main 2022 · 26 June · evening · Single correct · +4 / −1

The normal to the hyperbola
x²/a² - y²/9 = 1 at the point ( 8,3√3 ) on it passes through the point :

  1. (a) ( 15, - 2√3 )
  2. (b) ( 9,2√3 )
  3. (c) ( - 1,9√3 )
  4. (d) ( - 1,6√3 )
Answer and solution

Answer (c) ( - 1,9√3 )

Solution

Maths sir · 3 steps

First use the point to find a², then write the normal and test each option in it.

  1. Find a squared. The point (8, 3√3) lies on the hyperbola, so 64/a² - 27/9 = 1. That gives 64/a² = 4, so a² = 16 and b² = 9.
  2. Normal at a point. The normal to x²/a² - y²/b² = 1 at (x₁, y₁) is (a² x)/x₁ + (b² y)/y₁ = a² + b². Put in a² = 16, b² = 9, x₁ = 8 and y₁ = 3√3.

    16x/8 + 9y/3√3 = 25 ⇒ 2x + √3 y = 25

    OptionPoint2x + √3 y
    (a)(15, -2√3)30 - 6 = 24
    (b)(9, 2√3)18 + 6 = 24
    (c)(-1, 9√3)-2 + 27 = 25
    (d)(-1, 6√3)-2 + 18 = 16
  3. So. Only (-1, 9√3) satisfies 2x + √3 y = 25. The normal passes through option (c).

The idea to keep. If a constant in the curve is unknown, the given point fixes it first. Then the normal a²x/x₁ + b²y/y₁ = a² + b² comes out in one line.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Hyperbola

Asked about 60 times in JEE Main since 2020. This question is from 2022.