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

Question

Accepted Answer

Statement-I is true; Statement-II is true; Statement-II is not a correct explanation for Statement-I.. Statement-I is true because the line given is indeed a common tangent to both the circle and the parabola. Statement-II is also true as it correctly describes the condition for the slope of the tangent line, but it does not directly explain why the specific line in Statement-I is a common tangent.

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

Solution

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