# The points when the ellipse crosses the x -axis and to show that the tangent line at these points are parallel.

### 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 49E
To determine

## To find: The points when the ellipse crosses the x-axis and to show that the tangent line at these points are parallel.

Expert Solution

The ellipse crosses the x-axis at (3,0) and (3,0) and its tangent lines are parallel.

### Explanation of Solution

Given:

The equation of the ellipse x2xy+y2=3.

Derivative rules:

Product rule:ddx(fg)=fddx(g)+gddx(f)

Chain rule: dydx=dydududx

Calculation:

Obtain the points if the ellipse crosses the x-axis.

The ellipse crosses the x-axis. That is, y=0.

Substitute y=0 in the equation x2xy+y2=3,

x20+0=3x2=3x=±3

Therefore, the ellipse crosses the x-axis at (3,0) and (3,0).

Obtain the slope of the tangent at the points (3,0) and (3,0).

Differentiate x2xy+y2=3 implicitly with respect to x.

ddx(x2xy+y2)=ddx(3)ddx(x2)ddx(xy)+ddx(y2)=02xddx(xy)+ddx(y2)=0

Apply the product rule and the chain rule.

2xddx(xy)+ddy(y2)dydx=02x[xddx(y)+yddx(x)]+2ydydx=02x(xdydx+y)+2ydydx=02xy+dydx(2yx)=0

Separate dydx on one side of the equation,

dydx(2yx)=y2xdydx=y2x2yx

Thus, the derivative of the equation is dydx=y2x2yx.

That is, the slope of the tangent is dydx=y2x2yx.

The slope of the tangent at (3,0) is computed as follows,

dydx=02(3)0+3=233=2

Thus, the slope of the tangent at (3,0) is 2.

The slope of the tangent at (3,0) is computed as follows,

dydx=02(3)03=2(3)3=2

Thus, the slope of the tangent at (3,0) is 2.

Note that, the slope of the tangent lines at the points (3,0) and (3,0) are same.

Hence, the tangent lines at (3,0) and (3,0) are parallel to the equation is proved.

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