Let the tangent and normal at the point (3 √3, 1) on the ellipse x²/36+y²/4=1 meet the y-axis at the points A and B respectively. Let the circle C be drawn taking A B as a diameter and the line x=2 √5 intersect C at the points P and Q. If the tangents at the points P and Q on the circle intersect at the point (α, β), then α²-β² is equal to :

Solution

Maths sir · 5 steps

Line x = 2√5 is the chord of contact from (α, β), so match it with T = 0.

1. Find A and B. The tangent at (3√3, 1) is (3√3 x)/36 + y/4 = 1. Putting x = 0 gives A = (0, 4). The normal is 36x/3√3 - 4y/1 = 36 - 4. Putting x = 0 gives -4y = 32, so B = (0, -8).
2. Circle on AB. The centre is the midpoint (0, -2) and the radius is 12/2 = 6. So C is x² + (y+2)² = 36, that is x² + y² + 4y - 32 = 0.
3. Chord of contact. PQ joins the two points of contact of the tangents from (α, β), so its equation is T = 0 for that point.

   α x + β y + 2(y + β) - 32 = 0 ⇒ α x + (β + 2)y + 2β - 32 = 0

4. Compare with the line. This must be the line x = 2√5, which has no y term. So β + 2 = 0, giving β = -2. The equation becomes α x = 36, so α = 36/2√5 = 18/√5.
5. So. α² - β² = 324/5 - 4 = 304/5, which is option (b).

The idea to keep. When tangents at the ends of a chord meet at a point, that chord is the chord of contact T = 0 of the point. Compare coefficients with the given line to read off the point.

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