Answer
a) $(1,-1)$
b)
Work Step by Step
a)
$x^2-xy+y^2=3\\
2x-(xy'+y)+2yy'=0\\
2x-xy'-y+2yy'=0\\
y'(2y-x)=y-2x\\
y'=\frac{y-2x}{2y-x}$
Plug in $(-1,1)$ into $y′$ to find the gradient
$y'=\frac{(1)-2(-1)}{2(1)-(-1)}\\
y'=1$
To find the normal at some point we use the formula $y'\times m_{normal}=-1$
$1\times m_{normal}=-1\\
m_{normal}=1$
Then we find the equation of the normal
$y-1=-1(x-(-1))\\
y=-x$
Then we equate the normal line and the ellipse to find where they intersect.
$x^2-(x(-x))+(-x)^2=3\\
3x^2=3\\
x=1\Longrightarrow y=-1$
Therefore, the ellipse and the normal line intercept at $(1,-1)$
b)
