The parabolas : a x²+2 b x+c y=0 and d x²+2 e x+f y=0 intersect on the line y=1. If a, b, c, d, e, f are positive real numbers and a, b, c are in G.P., then :

Solution

Maths sir · 4 steps

Put y = 1 in both: the first gives a perfect square, so its only root must also satisfy the second.

1. Common root. At an intersection point on y = 1, the x-value satisfies both ax² + 2bx + c = 0 and dx² + 2ex + f = 0. So these two quadratics share a root.
2. Use the G.P. Since a, b, c are in G.P., b² = ac. The discriminant of the first quadratic is 4b² - 4ac = 0, so it has equal roots, x = -b/a. That root is the only candidate for the common root.
3. Substitute the root. Put x = -b/a into the second quadratic: db²/a² - 2eb/a + f = 0. Replace b² by ac and divide every term by c.

   d/a - 2eb/ac + f/c = 0 ⇒ d/a + f/c = 2e/b (using ac = b²)

4. So. The middle term is the average of the other two, so d/a, e/b, f/c are in A.P., which is option (a).

The idea to keep. When b² = ac, the quadratic ax² + 2bx + c is a perfect square with root -b/a. A common-root question then becomes plugging that single root into the other equation.

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