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AIEEE 2011

JEE Main 2011 Mathematics question on Conic Sections

Let P(6,3)P(6, 3) be a point on the hyperbola x2a2−y2b2=1\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1. If the normal at the point PP intersects the XX-axis at (9,0)(9, 0), then the eccentricity of the hyperbola is
  1. A.
    52\sqrt{\dfrac{5}{2}}
  2. B.
    32\sqrt{\dfrac{3}{2}}
  3. C.
    2\sqrt{2}
  4. D.
    3\sqrt{3}
Show answer and solution

Correct option: B

Solution

The normal to the hyperbola at point (x₁, y₁) is: a²x/x₁ + b²y/y₁ = a² + b².

At P(6,3), the normal is: a²x/6 + b²y/3 = a² + b².

It passes through (9, 0): 9a²/6 = a² + b², so (3/2)a² = a² + b², giving b² = a²/2.

Then e² = 1 + b²/a² = 1 + 1/2 = 3/2, so e = √(3/2).

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