If the line x-1=0 is a directrix of the hyperbola k x²-y²=6, then the hyperbola passes through the point :

Solution

Maths sir · 4 steps

Turn the directrix into an equation in a alone, solve it for a², and k follows straight away.

1. Standard form. Divide by 6 to get x²/(6/k) - y²/6 = 1. So a² = 6/k and b² = 6.
2. Use the directrix. The directrix is x = a/e, and it is given as x = 1, so a = e and a² = e². For a hyperbola, e² = 1 + b²/a² = 1 + 6/a².

   a² = 1 + 6/a² ⇒ a⁴ - a² - 6 = 0 ⇒ (a² - 3)(a² + 2) = 0

3. Find k. Since a² > 0, a² = 3. Then 6/k = 3 gives k = 2, and the hyperbola is 2x² - y² = 6. Check: e = √3 and a = √3, so a/e = 1.

   Point	2x² - y²	Equals 6?
   (a) (-2√5, 6)	40 - 36 = 4	No
   (b) (-√5, 3)	10 - 9 = 1	No
   (c) (√5, -2)	10 - 4 = 6	Yes
   (d) (2√5, 3√6)	40 - 54 = -14	No

4. So. Only (√5, -2) satisfies 2x² - y² = 6, so the answer is (c).

The idea to keep. A directrix x = a/e plus e² = 1 + b²/a² gives one equation in a². Solve it, then substitute each option into the curve.

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