Answer
$\dfrac{xx_0}{a^2}+\dfrac{yy_0}{b^2}-\dfrac{zz_0}{c^2}=1$
Work Step by Step
Formula to calculate tangent plane equation is:
$(x_2-x_1)f_x(x_1,y_1,z_1)+(y_2-y_1)f_y(x_1,y_1,z_1)+(z_2-z_1)f_x(x_1,y_1,z_1)=0$
At point$(x_0,y_0,z_0)$
$(x-x_0)(\dfrac{2x_0}{a^2})+(y-y_0)(\dfrac{2y_0}{b^2})+(z-z_0)(\dfrac{-2z_0}{c^2})=0$
This implies,
$\dfrac{2x_0}{a^2}(x)+\dfrac{2y_0}{b^2}(y)+\dfrac{2z_0}{c^2}(z)-[\dfrac{2x_0^2}{a^2}+\dfrac{2y_0^2}{b^2}+\dfrac{2z_0^2}{c^2}]=0$
Since, $\dfrac{x_0^2}{a^2}+\dfrac{y_0^2}{b^2}+\dfrac{z_0^2}{c^2}=1$
Hence,
$\dfrac{xx_0}{a^2}+\dfrac{yy_0}{b^2}-\dfrac{zz_0}{c^2}=1$
