JEE Main 2020 · 4 September · morning · Single correct · +4 / −1

Let P(3, 3) be a point on the hyperbola,
x²/a² - y²/b² = 1. If the normal to it at P intersects the x-axis at (9, 0) and e is its eccentricity, then the ordered pair (a², e²) is equal to :

  1. (a) ( 9/2,2 )
  2. (b) ( 3/2,2 )
  3. (c) (9,3)
  4. (d) ( 9/2,3 )
Answer and solution

Answer (d) ( 9/2,3 )

Solution

Maths sir · 4 steps

The normal through (3, 3) and (9, 0) has slope -1/2, and that ties b² to a².

  1. Slope of tangent. Differentiate x²/a² - y²/b² = 1 to get dy/dx = (b² x)/(a² y). At P(3, 3) the tangent slope is b²/a², so the normal slope is -a²/b².
  2. Slope of normal. The normal passes through (3, 3) and (9, 0), so its slope is (0 - 3)/(9 - 3) = -1/2. Then a²/b² = 1/2, which means b² = 2a².

    e² = 1 + b²/a² = 1 + 2 = 3

  3. Use the point. P lies on the curve, so 9/a² - 9/2a² = 1. This gives 9/2a² = 1, so a² = 9/2 and b² = 9.
  4. So. (a², e²) = (9/2, 3), which is option (d).

The idea to keep. For x²/a² - y²/b² = 1, the normal at (x₁, y₁) meets the x-axis at (x₁ e², 0). Here 9 = 3e² gives e² = 3 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 2020.