#### 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.

#### Answer

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

#### Explanation

**Given:**

The equation of the ellipse x2−xy+y2=3.

**Derivative rules:**

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

*Chain rule*: dydx=dydu⋅dudx

**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 x2−xy+y2=3,

x2−0+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 x2−xy+y2=3 implicitly with respect to *x*.

ddx(x2−xy+y2)=ddx(3)ddx(x2)−ddx(xy)+ddx(y2)=02x−ddx(xy)+ddx(y2)=0

Apply the product rule and the chain rule.

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

Separate dydx on one side of the equation,

dydx(2y−x)=y−2xdydx=y−2x2y−x

Thus, the derivative of the equation is dydx=y−2x2y−x.

That is, the slope of the tangent is dydx=y−2x2y−x .

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

dydx=0−2(−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=0−2(3)0−3=−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.