Hello, mathnerd11!

I got some strange answers, so I back-tracked.Given two curves: .y² .= .3 - 2x² .and .y² .= .x

Determine if they are orthogonal

(perpendicular at their points of intersection)

The first equation is: .2x² + y² .= .3

. . This is an ellipse centered at the origin.

The second is: .y² = x

. . This is a parabola opening to the right, vertex at the origin.

They will intersect in quadrants 1 and 4 (only).

To find the points of intersection, equate the functions and solve.

Since both equal y², we have: .x .= .3 - 2x²

. . We have a quadratic: .2x² + x - 3 .= .0

. . which factors: .(2x + 3)(x - 1) .= .0

. . and has roots: .x .= .-3/2, 1

But x = -3/2 is an extraneous root.

Hence, the graphs intersect at: .x = 1, .y = ±1

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Slope of the ellipse

Differentiate implicity: .2y·y' .= .-4x . . → . . me .= .-2x/y

Slope of the parabola

Differentiate implicitly: .2y·y' .= .1 . . → . . mp .= .1/(2y)

At (1, 1):

. . me .= .-2(1)/1 .= .-2

. . mp .= .1/(2·1) .= .1/2 . . . the slopes are perpendicular.

At (1,-1):

. . me .= .-2(1)/(-1) .= .2

. . mp .= .1/(2·-1) .= .-1/2 . . . the slopes are perpendicular

Therefore, the curves are orthogonal.