   Chapter 3.5, Problem 76E ### 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

# Find equations of both the tangent lines to the ellipse x2 + 4y2 = 36 that pass through the point (12, 3).

To determine

To find: The equation of the both tangent lines to the ellipse passing through the point.

Explanation

Given:

The equation of the ellipse x2+4y2=36.

The point is (12,3).

Derivative rules:

Chain rule: dydx=dydududx.

Calculation:

Obtain the slope of tangent to the equation.

Differentiate x2+4y2=36 implicitly with respect to x.

ddx(x2+4y2)=ddx(36)ddx(x2)+4ddx(y2)=02x+4ddx(y2)=0

Apply the chain rule and simplify the terms.

2x+4[ddy(y2)dydx]=02x+4[2ydydx]=02x+8ydydx=0dydx=x4y

Thus, the derivative of the equation of the ellipse is dydx=x4y.

That is, the slope of the tangent is dydx=x4y.

Let (a,b) be a point on the curve.

The slope of tangent to the curve at (a,b) is dydx=a4b.

The equation of the tangent line passing through the point (12,3) and the slope a4b is computed as follows.

y3=a4b(x12)y3=ax4b+12a4b4by12b=ax+12a

ax+4by=12a+12b (1)

Here, the tangent line is also passing through the point (a,b).

Substitute (a,b) for (x,y),

a(a)+4b(b)=12a+12b

a2+4b2=12(a+b) (2)

The value of the curve x2+4y2=36 at (a,b) is a2+4b2=36 (3)

Substitute the equation (3) in equation (2)

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