#### To determine

**To find:** The point such that the normal line intersect the ellipse at second time and to sketch the equation of the ellipse and the normal line.

#### Answer

The normal line is intersect to the ellipse second time at the point (1,−1).

#### Explanation

**Given:**

The equation of the ellipse is x2−xy+y2=3 and the point is (−1,1).

**Derivative rules:**

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

*Chain rule:* dydx=dydu⋅dudx

**Calculation:**

Obtain the slope of the normal line to the ellipse.

Differentiate x2−xy+y2=3 implicitly with respect to *x*,

ddx(x2−xy+y2)=ddx(3)ddx(x2)−ddx(xy)+ddx(y2)=ddx(3)2x−ddx(xy)+ddx(y2)=0

Apply the product rule and the chain rule,

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

Combine the dydx on one side of the equation,

(2y−x)dydx=y−2xdydx=y−2x2y−x

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

The slope of the tangent to the ellipse at (−1,1) is computed as follows,

m=y−2x2y−x

Substitute (−1,1) for (x,y),

m=1−2(−1)2(1)−(−1)=1+22+1=1

Thus, the slope of the tangent to the ellipse at (−1,1) is *m =* 1.

Note that, if the slope of the tangent line at (x1,y1) is m then the slope of the normal line at (x1,y1) is −1m.

Thus, the slope of the normal line to the ellipse at (−1,1) is −1.

Obtain the equation of the normal at (−1,1) and slope −1.

y−1=−1(x+1)y−1=−x−1y=−x

It is required that the normal line is intersect to the ellipse at second time.

Substitute y=−x in x2−xy+y2=3 to obtain the points.

x2−x(−x)+(−x)2=3x2+x2+x2=33x2=3x=±1

Substitute x=±1 in y=−x and obtain the values y=±1.

Thus, the points (−1,1) and (1,−1).

Therefore, the normal line is intersect to the ellipse at second time at (1,−1).

**Graph:**

Use online graphing calculator to sketch the equation of the ellipse and the normal line as shown in Figure 1.

From Figure 1, it is observed that the normal line y=−x to the ellipse at the point (−1,1) and it also intersect second time to the ellipse at the point (1,−1).