Why calculus method fails here for finding common tangent?

Question

Q. Find equation of common tangent(s) to the parabola $y^2 = 16\sqrt{3}x$ and the ellipse $2x^2 + y^2 = 4$ . When I solve this question by using methods of conic section like using equation of tangent at parabola $y = mx + \frac{a}{m}$ and for ellipse $y = mx \pm \sqrt{m^2 a^2 + b^2}$. and then comparing the constant term i get $m=2,-2$. But when I solve this question by calculus method by equating $\frac{dy}{dx}$ at point $(h,k)$ for both curves .I didnt get the right ans. This questions was asked by IIT JEE in 2012. Please Help...

Answer 1

A common tangent has a tangent point in the second quadrant for the ellipse and the first quadrant for the parabola. Assuming your slope of $2$ is correct, I get the point on the ellipse as $(\frac{-2\sqrt3}{3},\frac{2\sqrt3}{3})$ where $\frac{dy}{dx} = \frac{-2x}{\sqrt{4-2x^2}}$ and then the equation of the tangent $y = 2x + 2\sqrt3$. This works and is tangent to the parabola at $(\sqrt 3, 4\sqrt 3)$ where $\frac{dy}{dx} = \frac{2(3)^{1/4}}{\sqrt x} = 2$ at $x = \sqrt 3$.

Your mistake "may" have been in setting the derivatives equal to each other which gives a single x value which is not at either tangent point.