Answer
1) The point $(-1,3)$ is on the curve.
2) The tangent line to the curve at $(-1,3)$ is $$y=3x+6$$
3) The normal line to the curve at $(-1,3)$ is $$y=-\frac{1}{3}x+\frac{8}{3}$$
Work Step by Step
$$x^2y^2=9$$
1) Verify that the point $(-1,3)$ is on the curve
We substitute the coordinates of the point $(-1,3)$ into the formula of the function to see if it equals $9$ or not:
$$(-1)^2\times3^2=1\times9=9$$
Since it equals $9$, the point is on the curve.
2) Find the tangent at $(-1,3)$:
- Find the derivative of the function with implicit differentiation:
$$(2x)y^2+x^2(2yy')=0$$ $$2xy^2+2x^2yy'=0$$ $$xy^2+x^2yy'=0$$ $$x^2yy'=-xy^2$$ $$y'=-\frac{xy^2}{x^2y}=-\frac{y}{x}$$
- The slope of the tangent line at $(-1,3)$ is $$y'=-\frac{3}{-1}=3$$
- The tangent line to the given curve at $(-1,3)$ is
$$y-3=3(x+1)$$ $$y-3=3x+3$$ $$y=3x+6$$
3) Find the normal line at $(-1,3)$:
The normal line at $(-1,3)$ would be perpendicular with the tangent line at $(-1,3)$, so the product of the slopes of these two lines equals $-1$.
So if we call the slope of the normal line at $(-1,3)$ $k$, we would have
$$k\times3=-1$$ $$k=-\frac{1}{3}$$
- The normal line to the given curve at $(-1,3)$ therefore is
$$y-3=-\frac{1}{3}(x+1)$$ $$y-3=-\frac{1}{3}x-\frac{1}{3}$$ $$y=-\frac{1}{3}x+\frac{8}{3}$$
