Question

Show that the plane 2x-4y-z+3 = 0 touches the paraboloids x2 - 2y2=3z, find the point of contact.

Answer 1

If the plane touches the given parabola, then it must have at least one point of contact.
So, we will try to find point of contact first.

1) We have,
$2x - 4y - z + 3 = 0 \\ {x}^{2} - 2 {y}^{2} = 3z$
Put value of z from plane in Paraboloids,

Then,
${x}^{2} - 2 {y}^{2} = 3(2x - 4y + 3) \\ = > {x}^{2} - 6x - 2 {y}^{2} + 12y - 9 = 0 \\ - - - (1)$
If both the graph touches each other, then Discriminant of this quadratic in x must be 0 .

=>
${6}^{2} - 4 \times 1 \times ( - 2 {y}^{2} + 12y - 9) = 0 \\ = > 9 - (12y - 2 {y}^{2} - 9) = 0 \\ = > {y}^{2} - 6y + 9 = 0 \\ = > {(y - 3)}^{2} = 0 \\ = > y = 3$

2) Substitute value of y in equation. (1) ,

${x}^{2} - 6x + 12(3) - 2( {3}^{2} ) - 9 = 0 \\ = > {x}^{2} - 6x + 9 = 0 \\ = > {(x - 3)}^{2} = 0 \\ = > x = 3$

3) Substitute value of x & y in equation. of plane.
$2(3) - 4(3) - z + 3 = 0 \\ = > z = - 3$
Hence, Point of contact : (3,3,-3)

We got one point of contact of two surfaces,
=> Both curves touch each other at one point.