Prove that y^2=4x+4 and y^2=4-4x intersect at right angles.

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neela | High School Teacher | (Level 3) Valedictorian

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The 2 curves intersect at the poin whose x coordinate is the solution of:

4x+4=4-4x. Or

8x=0. Or x = 0.

So the  y coordinate  is given by y^2 = 4x+4 . Or y = 2 or y = -2. So the point of intersection of the curves are: (0,2 ) and (0,-2).

Therefore the slope of  y^2=4x at (0,2) and (0-2) are  given by dy/dx at these points (0,2) and (0, -2)

Diferentiating y^2 = 4x both sides with respect to x, we get:

2y dy/dx = 4. Or  (dy/dx) = 4/(2y) = (2/y ) = 1 at (0,2) and dy/dx = m1 = 4/(-4) = -1 at (0,-2).

Similarly for the curve y^2=4-4x we get dy/ dx = -4/y = -4/(2y) m2 =-1 at (0,2) and dy/dx = -4/(-2*-2) = 1 at (0,-2)

Thus the product of the slopes m1*m2  = -1of the tangents for both points of interesection. So the tagents should intersect at right angles.

 

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

To find out the intersection point, we have to put the 2 equations into the relation:

4x+4 = 4-4x

For x=0, we'll have y^2 = 4*0 + 4

y^2 = 4

y1=2 and y2=-2

To verify if the 2 curves (parabolas) are perpendicular, we have to verify if the product of the slopes of their tangent lines, in their intersection point, is -1.

First, let's calculate their slopes. In order to do so, we'll calculate their derivatives.

The derivative, with respect to x, for the first parabola:

2y*y' = 4

y' = 4/2y

y' = 2/y, the slope of the first parabola.

The derivative, with respect to x, for the second parabola:

2y*y' = -4

y' = -2/y, the slope of the second parabola.

Now, we'll check if the product of slopes is -1.

(2/y)(-2/y) = -4/y^2

We'll substitute y^2 by the value found at the point of intersection, y^2=4

-4/y^2 = -4/4=-1

The product of the slopes yields -1, so the parabolas are intersecting eachother at right angle.

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