Tangent Planes and Linear Approximations — Question 7

Question 7

For the paraboloid $$z=x^2+2y^2,$$ find the tangent plane parallel to the plane $$2x-4y+z=7.$$

Tasks

1. Determine the point of tangency.

2. Write the tangent plane in standard form.

3. Find the perpendicular distance between the two parallel planes.

Question 7 – Solution

Strategy. Match the $x$- and $y$-slope coefficients of the tangent plane to those of the given plane.

Step 1: Match slopes The given plane is $z=7-2x+4y$, so its slopes are $-2$ and $4$. For $f=x^2+2y^2$, $$f_x=2x,\qquad f_y=4y.$$ Thus $2x=-2$ and $4y=4$, giving $(x,y)=(-1,1)$. The height is $z=1+2=3$.

Step 2: Tangent plane At $(-1,1,3)$, $$z-3=-2(x+1)+4(y-1),$$ or $$\boxed{2x-4y+z=-3}.$$

Step 3: Separation For parallel planes $2x-4y+z=7$ and $2x-4y+z=-3$, the distance is $$\boxed{\frac{|7-(-3)|}{\sqrt{2^2+(-4)^2+1^2}} =\frac{10}{\sqrt{21}}}.$$ The identical left sides verify parallelism, and substituting $(-1,1,3)$ into the tangent plane gives $-3$.