Answer
$y=mx \pm \sqrt{a^2m^2+b^2}$
Work Step by Step
The standard equation of the ellipse is: $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ ...(1)
The equation of the tangent with slope $m$ can be written as: $y=mx+c$
Thus, equation (1) becomes:
$\dfrac{x^2}{a^2}+\dfrac{(mx+c)^2}{b^2}=1$
This gives: $(\dfrac{1}{a^2}+\dfrac{m^2}{b^2})x^2+\dfrac{2mcx}{b^2}+(\dfrac{c^2}{b^2}-1)=0$
This forms a quadratic equation in the variable of $x$.
Thus, $D=b^2-4ac=0 \implies (\dfrac{2mc}{b^2})^2+4(\dfrac{1}{a^2}+\dfrac{m^2}{b^2})(\dfrac{c^2}{b^2}-1)=0$
or, $(\dfrac{m^2}{b^4})c^2-(\dfrac{1}{a^2b^2}+\dfrac{m^2}{b^4})c^2+(\dfrac{1}{a^2}-\dfrac{m^2}{b^2})=0$
or, $-(\dfrac{1}{a^2b^2})c^2=-(\dfrac{1}{a^2}-\dfrac{m^2}{b^2})$
This gives: $c^2=b^2+a^2m^2$
or, $c=\pm \sqrt{b^2+a^2m^2}$
Hence, the equation of the tangent with slope $m$ is:
$y=mx \pm \sqrt{a^2m^2+b^2}$
Hence, the result has been proved.
