# The ellipse and the hyperbola are orthogonal trajectories if A 2 &lt; a 2 and a 2 − b 2 = A 2 + B 2 . ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805 ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 3.5, Problem 45E
To determine

## To show: The ellipse and the hyperbola are orthogonal trajectories if A2<a2 and a2−b2=A2+B2.

Expert Solution

### Explanation of Solution

Given:

The equation of the ellipse x2a2+y2b2=1 (1)

The equation of the hyperbola x2A2y2B2=1 (2)

Derivative rules:

Chain rule: dydx=dydududx

Proof:

Consider equation of the ellipse x2a2+y2b2=1.

Differentiate implicitly with respect to x,

ddx(x2a2+y2b2)=ddx(1)ddx(x2a2)+ddx(y2b2)=01a2ddx(x2)+1b2ddx(y2)=02xa2+1b2ddx(y2)=0

Apply the chain rule and simplify the terms,

2xa2+1b2ddy(y2)dydx=02xa2+2yb2dydx=02yb2dydx=2xa2dydx=xb2ya2

Thus, the derivative of the equation of the ellipse is dydx=xb2ya2.

That is, the slope of the tangent to equation of the ellipse is m1=xb2ya2.

Consider equation of the ellipse x2A2y2B2=1.

Differentiate implicitly with respect to x,

ddx(x2A2y2B2)=ddx(1)ddx(x2A2)ddx(y2B2)=01A2ddx(x2)1B2ddx(y2)=02xA21B2ddx(y2)=0

Apply the chain rule and simplify the terms,

2xA21B2ddy(y2)dydx=02xA22yB2dydx=02yB2dydx=2xA2dydx=xB2yA2

Thus, the derivative of the equation of the ellipse is dydx=xB2yA2.

That is, the slope of the tangent to equation of the hyperbola is m2=xB2yA2.

Suppose that A2<a2 and a2b2=A2+B2.

It is required to prove that the equations are orthogonal trajectories.

That is, the product of their slope is -1.

m1m2=xb2ya2xB2yA2=b2B2a2A2(x2y2)

m1m2=b2B2a2A2(x2y2) (3)

Subtract the equation (2) from the equation (1),

x2a2+y2b2x2A2+y2B2=0y2b2+y2B2=x2A2x2a2B2y2+y2b2b2B2=a2x2x2A2a2A2

y2(b2+B2)b2B2=x2(a2A2)a2A2 (4)

Since a2b2=A2+B2, a2A2=b2+B2.

Substitute a2A2=b2+B2 in equation (4),

y2(b2+B2)b2B2=x2(b2+B2)a2A2y2b2B2=x2a2A2x2y2=a2A2b2B2

Substitute the value x2y2=a2A2b2B2 in equation (3),

m1m2=b2B2a2A2(a2A2b2B2)=1

Therefore, the ellipse and the hyperbola are orthogonal trajectories if a2b2=A2+B2 and A2<a2.

Hence the required result is proved.

### Have a homework question?

Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers! 