A hyperbola having the transverse axis of length √2 has the same foci as that of the ellipse 3x² + 4y² = 12, then this hyperbola does not pass through which of the following points?

Solution

Maths sir · 3 steps

Take the foci from the ellipse, build the hyperbola, then plug each point in and see which one fails.

1. Foci of ellipse. Divide by 12 to get x²/4 + y²/3 = 1. For an ellipse c² = a² - b² = 4 - 3 = 1, so the foci are (± 1, 0).
2. Build the hyperbola. The hyperbola has the same foci, so its centre is the origin, its transverse axis lies along the x-axis, and c = 1. The transverse axis is 2a = √2, so a² = 1/2. For a hyperbola c² = a² + b², so b² = 1 - 1/2 = 1/2.

   x²/(1/2) - y²/(1/2) = 1 ⇒ x² - y² = 1/2

   Point	x² - y²	On the hyperbola?
   (a) (1, -1/√2)	1 - 1/2 = 1/2	yes
   (b) (√(3/2), 1/√2)	3/2 - 1/2 = 1	no
   (c) (-√(3/2), 1)	3/2 - 1 = 1/2	yes
   (d) (1/√2, 0)	1/2 - 0 = 1/2	yes

3. So. Only the point in (b) gives x² - y² = 1 instead of 1/2, so the hyperbola does not pass through it.

The idea to keep. Ellipse: c² = a² - b². Hyperbola: c² = a² + b². Shared foci means the same c. Find c from one curve and use it to build the other.

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