Question:
Find the equations of the common tangents to the parabola $y^2=2ax$ and $x^2= 2by$.
So, as we know, the equation of the tangent of $y^2=2ax$ in slope form is $y=mx+\frac{a}{2m}$; and the equation of tangent of $x^2=2by$ in slope form is $y=mx-\frac{bm^2}{2}$. Since here we are talking about the same tangent, therefore the '$m$' in both the equations is equal. So then when i just compare the two equations, the answer i get is not correct. So, can someone please point out what i have done wrong here?
$m = - \sqrt[3]{\frac{a}{b}} \ $ is correct. The official solution is not correct for the stated question.
Tangent to $y^2 = 2 a x$ can be written as $y = m x + \frac{a}{2m}$
From here, we can either proceed the way you did or we can plug this in $x^2 = 2 b y$.
$x^2 = 2b (mx + \frac{a}{2m}) \implies (x-bm)^2 - b^2m^2 - \frac{ab}{m} = 0$
As the line is tangent to the parabola, we have only one solution to the equation.
So we must have, $b^2m^2 + \frac{ab}{m} = 0 \implies m = - \sqrt[3]{\frac{a}{b}} \ , $ which is same as your answer.