# The equation of the both tangent lines to the ellipse passing through the point.

### 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 52E
To determine

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

Expert Solution

The equation of the tangent line to the curve passing through the point (12,3) at (0,3) and (245,95) are y=3 and y=23x5.

### Explanation of Solution

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).

12(a+b)=36a+b=3b=3a

Substitute b=3a in equation (3).

a2+4(3a)2=36a2+4(9+a26a)=36a2+36+4a224a36=05a224a=0

a(5a24)=0a=0 or a=245

Substitute a=0  in b=3a,

Thus, the value of b is 3.

Substitute a=245 in b=3a.

b=3245=15245=95

Thus, the points are (0,3) and (245,95).

Substitute (0,3) for (a,b) in equation (1),

(0)x+4y(3)=12(0)+12(3)12y=36y=3

Thus, the equation of the tangent line to the curve passing through the point (12,3) at (0,3) is y=3.

Substitute (245,95) for (a,b) in equation (1).

(245)x+4y(95)=12(245)+12(95)24x36y=28810824x36y=180y=23x5

Thus, the equation of the tangent line to the curve passing through the point (12,3) at (245,95) is y=23x5.

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