   Chapter 3.5, Problem 44E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Show by implicit differentiation that the tangent to the ellipse a 2 a 2 + y 2 b 2 = 1 at the point (x0, y0) is x 0 x a 2 + y 0 y b 2 = 1

To determine

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

Explanation

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

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