Answer
$y=mx \pm \sqrt{a^2m^2+b^2}$
Work Step by Step
As we know that the equation of tangent with slope $m$ is: $y=mx+c$
The standard form of the ellipse is given as:
$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$
$\implies \dfrac{x^2}{a^2}+\dfrac{(mx+c)^2}{b^2}=1$
$\implies (\dfrac{1}{a^2}+\dfrac{m^2}{b^2})x^2+\dfrac{2mcx}{b^2}+(\dfrac{c^2}{b^2}-1)=0$
This shows an quadratic equation with the variable of $x$.
Here, $Discriminant, D=b^2-4ac=0$
$(\dfrac{2mc}{b^2})^2+4(\dfrac{1}{a^2}+\dfrac{m^2}{b^2})(\dfrac{c^2}{b^2}-1)=0$
This implies that
$(\dfrac{m^2}{b^4})c^2-(\dfrac{1}{a^2b^2}+\dfrac{m^2}{b^4})c^2+(\dfrac{1}{a^2}-\dfrac{m^2}{b^2})=0$
or, $-(\dfrac{1}{a^2b^2})c^2=-(\dfrac{1}{a^2}-\dfrac{m^2}{b^2})$
Therefore, we have $c^2=a^2m^2+b^2$
This gives: $c=\pm \sqrt{b^2+a^2m^2}$
Hence, the result has been verified that
$y=mx \pm \sqrt{a^2m^2+b^2}$
