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Single Variable Calculus: Concepts...

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

Single Variable Calculus: Concepts...

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

Solutions

Chapter 3.5, Problem 40E
To determine

To show: The tangent line to the ellipse at the point (x0,y0) is x0xa2+y0yb2=1.

Expert Solution

Explanation of Solution

Given:

The equation of ellipse is x2a2+y2b2=1.

The equation of ellipse at (x0,y0) is x02a2+y02b2=1.

Derivative rule: Chain rule

If y=f(u) and u=g(x)  are both differentiable function, then dydx=dydududx.

Formula used:

The equation of the tangent line at (x1,y1) is, yy1=m(xx1) (1)

Where, m is the slope of the tangent line at (x1,y1) and m=dydx|(x1,y1).

Proof:

Obtain the equation of tangent line to the ellipse at (x0,y0).

Consider the equation of ellipse x2a2+y2b2=1.

Differentiate implicitly with respect to x,

ddx(x2a2+y2b2)=ddx(1)ddx(x2a2)+ddx(y2b2)=ddx(1)1a2ddx(x2)+1b2ddx(y2)=ddx(1)

Apply the chain rule (1) and simplify the terms,

1a2(2x)+1b2[ddy(y2)dydx]=01a2(2x)+1b2[2ydydx]=02xa2+2yb2dydx=02yb2dydx=2xa2

Multiply the equation by b22y,

dydx=2xa2×b22ydydx=xb2ya2

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

The slope of the tangent line at (x0,y0) is computed as follows,

Substitute (x0,y0) for (x,y) in dydx=xb2ya2,

m=dydx|(x0,y0)=x0b2y0a2

Therefore, the slope of the tangent at (x0,y0) is m=x0b2y0a2.

Substitute (x0,y0) for (x,y) and m=x0b2y0a2 in equation (1),

yy0=x0b2y0a2(xx0)(yy0)y0a2=x0b2(xx0)yy0a2y02a2=xx0b2+x02b2yy0a2+xx0b2=y02a2+x02b2

Divided the equation by a2b2 on both sides,

yy0a2a2b2+xx0b2a2b2=y02a2a2b2+x02b2a2b2yy0b2+xx0a2=y02b2+x02a2xx0a2+yy0b2=x02a2+y02b2

Substitute x02a2+y02b2=1 in the above equation,

xx0a2+yy0b2=1

Hence, it can be concluded that the tangent line to the ellipse at the point (x0,y0) is x0xa2+y0yb2=1.

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