Question

Given : A circle, \(2x^2 + 2y^2 = 5\) and a parabola, \(y^2 = 4\sqrt{5}\,x\).
Statement-1 : An equation of a common tangent to these curves is \(y = x + \sqrt{5}\).
Statement-2 : If the line, \(y = mx + \frac{\sqrt{5}}{m}\;(m \ne 0)\) is their common tangent, then \(m\) satisfies \(m^4 - 3m^2 + 2 = 0\).

• A. Statement-1 is true; Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
• B. Statement-1 is true; Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
• C. Statement-1 is true; Statement-2 is false.
• D. Statement-1 is false; Statement-2 is true.

Hint:

Find the common tangents: impose the circle tangency condition \(c^2 = r^2(1+m^2)\) on the parabola tangent \(y = mx + \frac{\sqrt{5}}{m}\). Then test whether the slopes found satisfy the quartic of Statement-2.

Answer 1

The Correct Option is B

Circle: \(x^2 + y^2 = \frac{5}{2}\), so \(r^2 = \frac{5}{2}\). Parabola: \(y^2 = 4ax\) with \(a = \sqrt{5}\), tangent \(y = mx + \frac{\sqrt{5}}{m}\).
This line touches the circle if \(c^2 = r^2(1 + m^2)\): \(\frac{5}{m^2} = \frac{5}{2}(1 + m^2) \Rightarrow m^4 + m^2 - 2 = 0 \Rightarrow (m^2 + 2)(m^2 - 1) = 0\), so \(m = \pm 1\).
Statement-1: \(m = 1\) gives \(y = x + \sqrt{5}\), a common tangent. So Statement-1 is true.
Statement-2: \(m^4 - 3m^2 + 2 = (m^2 - 1)(m^2 - 2)\). The slopes of the common tangents, \(m = \pm 1\), do satisfy this equation, so the statement "if the line is a common tangent then \(m\) satisfies \(m^4 - 3m^2 + 2 = 0\)" is true. But it is not the condition from which the common tangent is obtained (that is \(m^4 + m^2 - 2 = 0\)), so it does not explain Statement-1.
Answer: (2) Statement-1 is true; Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.