Answer
$\dfrac{2xx_0}{a^2}+\dfrac{2yy_0}{b^2}=\dfrac{z_0}{c}+\dfrac{z}{c}$
Work Step by Step
Our aim is to determine the tangent plane equation.
The general form is: $(x_2-x_1)f_x(x_1,y_1,z_1)+(y_2-y_1)f_y(x_1,y_1,z_1)+(z_2-z_1)f_x(x_1,y_1,z_1)=0$ ...(1)
From the given data, we have $f(x,y,z)=(x_0,y_0,z_0)$
Equation (1), becomes:
Thus,
$(x-x_0)(\dfrac{2x_0}{a^2})+(y-y_0)(\dfrac{2y_0}{b^2})+(z-z_0)(\dfrac{-1}{c})=0$
After simplifications, we get
$\dfrac{2x_0}{a^2}(x)+\dfrac{2y_0}{b^2}(y)+\dfrac{2z_0}{c^2}(z)+\dfrac{z_0}{c}-\dfrac{2y_0^2}{b^2}-\dfrac{z}{c}=0$
As we are given $\dfrac{x_0^2}{a^2}+\dfrac{y_0^2}{b^2}=\dfrac{z_0}{c}$
The desired answer is: $\dfrac{2xx_0}{a^2}+\dfrac{2yy_0}{b^2}=\dfrac{z_0}{c}+\dfrac{z}{c}$
