Answer
See solution
Work Step by Step
$~\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}~ = 1$
We differentiate on both sides :
$\frac{2x}{a^{2}} ~+~\frac{2y}{b^{2}}\times\frac{dy}{dx} = 0 $ $~~~~$ (Chain rule)
$\frac{dy}{dx}=-\frac{~xb^{2}~}{~ya^{2}~}$
The above equation is the gradient of the ellipse
$Substitute ~point ~(x_{0},y_{0})$ to the above equation, we will get the gradient of the tangent line at point $~(x_{0},y_{0})$ :
$\frac{dy}{dx}\vert$$_{({x_{0},{y_{0}}}_{)}}$$=$$~-\frac{~x_{0}~b^{2}~}{~y_{0}~a^{2}~}$
$~~~~\frac{y~-~y_{0}}{x~-~x_{0}} = -\frac{x_{0}b^{2}}{y_{0}a^{2}}$
Use cross product:
$~~~(y~-~y_{0} )\times(y_{0}a^{2})= -(x~-~x_{0} )\times(x_{0}b^{2}) $
We arrange the above equation will get :
$\frac{x~x_{0}}{a^{2}} + \frac{y~y_{0}}{b^{2}} = \frac{x_{0}^{2}}{a^{2}}+\frac{x_{0}^{2}}{b^{2}}$
The right side is equal to $1$ in accordance with the ellipse equation, hence :
$\frac{x~x_{0}}{a^{2}} + \frac{y~y_{0}}{b^{2}} =1$
