## Calculus: Early Transcendentals 8th Edition

a) $(1,-1)$ b)
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)