Let the lengths of the transverse and conjugate axes of a hyperbola in standard form be 2 a and 2 b, respectively, and one focus and the corresponding directrix of this hyperbola be (-5,0) and 5 x+9=0, respectively. If the product of the focal distances of a point (α, 2 √5) on the hyperbola is p, then 4 p is equal to ___________.

Solution

Maths sir · 4 steps

Focus gives ae, directrix gives a/e: multiply them and a² falls out at once.

1. Find a and e. The focus (-5, 0) gives ae = 5. The directrix x = -9/5 gives a/e = 9/5. Multiplying, a² = 9, so a = 3 and e = 5/3.
2. Find b. b² = a²(e² - 1) = 9(25/9 - 1) = 16. The hyperbola is x²/9 - y²/16 = 1.
3. Locate the point. Put y² = 20: α²/9 = 1 + 20/16 = 9/4, so α² = 81/4 and α = ±9/2.

   p = |eα + a| |eα - a| = e²α² - a² = 25/9×81/4 - 9 = 189/4

4. So. The focal distances are 21/2 and 9/2, so p = 189/4 and 4p = 189.

The common slip: 47.25. Stopping at p itself gives 189/4 = 47.25. The question asks for 4p.

The idea to keep. For a hyperbola, the focal distances of (x, y) are |ex + a| and |ex - a|, so their product is e²x² - a². A focus with its directrix gives ae and a/e together.

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