Answer
$y=mx \pm \sqrt{a^2m^2+b^2}$
Work Step by Step
The standard equation of the ellipse is: $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ ...(I)
The equation of the tangent with slope $m$ can be written as: $y=mx+c$ ...(II)
From the equation (1) , we have
$\dfrac{x^2}{a^2}+\dfrac{(mx+c)^2}{b^2}=1$
$\implies (\dfrac{1}{a^2}+\dfrac{m^2}{b^2})x^2+\dfrac{2mcx}{b^2}+(\dfrac{c^2}{b^2}-1)=0$
Let us find out the discriminant.
That is, $D=b^2-4ac=0$
or, $(\dfrac{2mc}{b^2})^2+4(\dfrac{1}{a^2}+\dfrac{m^2}{b^2})(\dfrac{c^2}{b^2}-1)=0$
Then, we have $-(\dfrac{1}{a^2b^2})c^2=-(\dfrac{1}{a^2}-\dfrac{m^2}{b^2})$
$\implies c^2=b^2+a^2m^2$
$\implies c=\pm \sqrt{a^2m^2+b^2``}$
Hence, we have found the equation of the tangent with slope $m$:
$y=mx \pm \sqrt{a^2m^2+b^2}$
