   Chapter 2.6, Problem 60E

Chapter
Section
Textbook Problem

# Find equations of both the tangent lines to the ellipse x 2 +   4 y 2 =   36 that pass through the point (12, 3).

To determine

To Find:

Equation of tangent lines that pass through the given point.

Explanation

1) Formula:

i. Sum or difference rule

ddxfx ±gx=ddx fx± ddxgx ;

ii. Derivative of constant

ddxc=0,

iii. Product rule

ddxfx*gx=fxddxgx+gxddx fx

iv. Power rule

ddxxn=n xn-1

2) Given:

x2+4y2=36,  point (12,3)

3) Calculation:

Differentiating given expression implicitly with respect to x on both sides,

ddxx2+4y2=ddx36

2x+8y dydx=0

Use of Product Rule, Implicit differentiation

8ydydx=-2x

dydx=-2x8y

dydx=-x4y Thus it follows that the tangent at a point (x1,y1) on the ellipse has slope dydx=m=-x14y1

By the of the equation of curve at (x1,y1) is we also have that

x12+4y12=36

Suppose that a tangent line passing through point (12,3) touches the ellipse at (x1,y1)  then from definition of slope a line and the above we have

3-y1=-x14y112-x13-y1= - 12x1-x124y1

Multiplying both side of equation by 4y1,

4y13-y1= - 12x1+x12

12y1-4y12= - 12x1+x12

Now, since the point (x1,y1) lies on the curve, it must satisfy  x12+4y12=36. Thus,x12=36-4y12

12y1-4y12= - 12x1+36-4y12

12y1= - 12x1+36

12y1+ 12x1=36

y1+x1=3

x1=3-y1

Substitute value of x1 in the equation of given curve we have,

3-y12+4y12=36

9-6y1+y12+4y12=

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