# Thread: Orthogonal Curves...

1. ## Orthogonal Curves...

Hey guys, I'm having trouble with a problem solving question i was assigned earlier this week. We are given two curves: 1: y^2 = 3-2x^2 and 2: y^2 = x, and we must first determine if they are orthogonal, and then determine at which points they intersect at (at 90 degrees of course). Any feedback would be wonderful

Thanks again,

Math Nerd

2. Hello, mathnerd11!

Given two curves: . .= .3 - 2x² .and . .= .x

Determine if they are orthogonal
(perpendicular at their points of intersection)
I got some strange answers, so I back-tracked.

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 . . . . m
e .= .-2x/y

Slope of the parabola
Differentiate implicitly: .2y·y' .= .1 . . . . m
p .= .1/(2y)

At (1, 1):
. . m
e .= .-2(1)/1 .= .-2
. . m
p .= .1/(2·1) .= .1/2 . . . the slopes are perpendicular.

At (1,-1):
. . m
e .= .-2(1)/(-1) .= .2
. . m
p .= .1/(2·-1) .= .-1/2 . . . the slopes are perpendicular

Therefore, the curves are orthogonal.