# Show that the ellipse 2x^2+y^2= 3 and the parabola y^2= x are orthogonal (perpendicular) at their points of intersection.

### 1 Answer | Add Yours

The ellipse 2x^2 + y^2 = 3 and the parabola y^2 = x intersect at the solution of 2x^2 + x = 3

=> 2x^2 + 3x - 2x - 3 = 0

=> x(2x + 3) - 1(2x + 3) = 0

=> x = 1 and x = -3/2

As y^2 = -3/2 has no real roots consider only y^2 = x = 1

The points of intersection are (1, 1) and (1, -1)

The normal of the curves at the point of intersection are perpendicular if the tangent is perpendicular as the normal is perpendicular to the tangent.

For any curve the slope of the tangent at any point is equal to the value of the derivative y' at that point.

For the ellipse 2x^2 + y^2 = 3

=> 4x + 2y*y' = 0

=> y' = -2x/y

For the parabola y' = x/2y

At the point (1,1) the product of the slope of the tangents is -1. The same is the case at (1, -1).

**This shows that the ellipse 2x^2+y^2= 3 and the parabola y^2= x are orthogonal (perpendicular) at their points of intersection.**