# Show that the equation of the tangent plane of the paraboloid z=(x^2)/(a^2)+(y^2)/(b^2) at point (x0,y0,z0) can be written as z+z0=(2x(x0))/(a^2)+(2y(y0))/(b^2).

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to remember that the equation of the tangent plane at a point `(x_0,y_0)` , on the surface `z = f(x,y)` , is:

`z = f(x_0,y_0) + f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0)`

You need to find the partial derivative `f_x` of the given function, such that:

`f_x = 2x/a^2 => f_x(x_0,y_0) = (2x_0)/a^2`

You need to find the partial derivative `f_y` of the given function, such that:

`f_y = 2y/b^2 => f_y(x_0,y_0) = (2y_0)/b^2`

You need to evaluate `f(x_0,y_0)` , such that:

`f(x_0,y_0) = (x_0)^2/a^2 + (y_0)^2/b^2`

`z =(x_0)^2/a^2 + (y_0)^2/b^2 + (2x_0)/a^2(x-x_0) + (2y_0)/b^2(y-y_0)`

Opening the brackets yields:

`z =(x_0)^2/a^2 + (y_0)^2/b^2 + (2x x_0)/a^2 - (2x^2_0)/a^2 + (2yy_0)/b^2 - (2y^2_0)/b^2`

`z_(x_0,y_0) = (2x x_0)/a^2 + (2yy_0)/b^2 - (x_0)^2/a^2 - (y_0)^2/b^2`

Hence, evaluating the equation of the tangent plane to the surface of paraboloid `z =(x)^2/a^2 + (y)^2/b^2 ` , at the point `(x_0,y_0)` , yields `z_(x_0,y_0) = (2x x_0)/a^2 + (2yy_0)/b^2 - (x_0)^2/a^2 - (y_0)^2/b^2` .