Given a circle 2x^2+2y^2=5 and a parabola y^2=4√5x.

Question

Given a circle $2x^2+2y^2=5$ and a parabola $y^2=4\sqrt5x$.
Statement 1: An equation of a common tangent to these curves is $y=x+\sqrt5$.
Statement 2: If the line $y=mx+\dfrac{\sqrt5}{m}\ (m
e0)$ is their common tangent, then $m$ satisfies $m^4-3m^2+2=0$.

Accepted Answer

Tangent to $y^2=4\sqrt5x$: $y=mx+\dfrac{\sqrt5}{m}$. For tangency to $2x^2+2y^2=5$ i.e. $x^2+y^2=\tfrac52$ (radius $\sqrt{5/2}$), the perpendicular distance from the centre equals the radius:
$$\frac{\sqrt5/m}{\sqrt{1+m^2}}=\sqrt{\frac52}\ \Rightarrow\ \frac{5}{m^2(1+m^2)}=\frac52\ \Rightarrow\ m^2(1+m^2)=2\ \Rightarrow\ m^4+m^2-2=0\ \Rightarrow\ (m^2+2)(m^2-1)=0\Rightarrow m=\pm1.$$

So Statement 1 ($m=1$ giving $y=x+\sqrt5$) is true. Testing $m=\pm1$ in Statement 2's equation $m^4-3m^2+2=0$: $1-3+2=0$ --- also satisfied, so Statement 2 is true numerically, but the *correct* governing equation is $m^4+m^2-2=0$, not $m^4-3m^2+2=0$; hence Statement 2, though true, is not the correct explanation of Statement 1.

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