Answer
1) The point $(-2,1)$ is on the curve.
2) The tangent line to the curve at $(-2,1)$ is $y=-x-1$
3) The normal line to the curve at $(-2,1)$ is $y=x+3$
Work Step by Step
$$y^2-2x-4y-1=0$$
1) Verify that the point $(-2,1)$ is on the curve
We substitute the coordinates of the point $(-2,1)$ into the formula of the function to see if it equals $0$ or not:
$$1^2-2(-2)-4\times1-1=1+4-4-1=0$$
Since it equals $0$, the point is on the curve.
2) Find the tangent at $(-2,1)$:
- Find the derivative of the function with implicit differentiation:
$$2yy'-2-4y'-0=0$$ $$2yy'-4y'=2$$ $$yy'-2y'=1$$ $$y'(y-2)=1$$ $$y'=\frac{1}{y-2}$$
- The slope of the tangent line at $(-2,1)$ is $$y'=\frac{1}{1-2}=-1$$
- The tangent line to the given curve at $(-2,1)$ is
$$y-1=-(x+2)$$ $$y-1=-x-2$$ $$y=-x-1$$
3) Find the normal line at $(-2,1)$:
The normal line at $(-2,1)$ would be perpendicular with the tangent line at $(-2,1)$, so the product of the slopes of these two lines equals $-1$.
So if we call the slope of the normal line at $(-2,1)$ $k$, we would have
$$k\times(-1)=-1$$ $$k=1$$
- The normal line to the given curve at $(-2,1)$ therefore is
$$y-1=x+2$$ $$y=x+3$$
