Let the hyperbola H: x²/a²-y²/b²=1 pass through the point (2 √2,-2 √2). A parabola is drawn whose focus is same as the focus of H with positive abscissa and the directrix of the parabola passes through the other focus of H. If the length of the latus rectum of the parabola is e times the length of the latus rectum of H, where e is the eccentricity of H, then which of the following points lies on the parabola?

Solution

Maths sir · 4 steps

Focus at ae, directrix through -ae: the gap is 2ae, so the parabola's latus rectum is 4ae.

1. Set up parabola. The foci of H are (± ae, 0). The parabola has focus (ae, 0) and directrix x = -ae, the line perpendicular to the axis through the other focus. The focus-to-directrix distance is 2ae, so the parabola's latus rectum is 2 × 2ae = 4ae.

   4ae = e · 2b²/a ⇒ b² = 2a²

2. Find a and e. The point (2√2, -2√2) is on H, so 8/a² - 8/b² = 1. Putting b² = 2a² gives 8/a² - 4/a² = 1, so a² = 4 and b² = 8. Then e² = 1 + 8/4 = 3, so e = √3 and ae = 2√3.
3. Parabola equation. The vertex is midway between the focus (2√3, 0) and the directrix x = -2√3, which is the origin. So the parabola is y² = 4(2√3)x = 8√3 x.

   Point	y²	8√3 x	On parabola?
   (a) (2√3, 3√2)	18	48	no
   (b) (3√3, -6√2)	72	72	yes
   (c) (√3, -√6)	6	24	no
   (d) (3√6, 6√2)	72	72√2	no

4. So. Only (3√3, -6√2) satisfies y² = 8√3 x, so the answer is (b).

The idea to keep. For a parabola, the latus rectum is twice the focus-to-directrix distance. Turn the geometric condition into that distance first, and the rest is just solving for a and b.

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