University Calculus: Early Transcendentals (3rd Edition)

1) The point $(2,3)$ is on the curve. 2) The tangent line to the curve at $(2,3)$ is $y=7/4x-1/2$ 3) The normal line to the curve at $(2,3)$ is $y=-4/7x+29/7$
$$x^2+xy-y^2=1$$ 1) Verify that the point $(2,3)$ is on the curve We substitute the coordinates of the point $(2,3)$ into the formula of the function to see if it equals $1$ or not: $$2^2+2\times3-3^2=4+6-9=1$$ Since it equals $1$, the point is on the curve. 2) Find the tangent at $(2,3)$: - Find the derivative of the function with implicit differentiation: $$2x+(y+xy')-2yy'=0$$ $$2x+y=2yy'-xy'$$ $$y'(2y-x)=2x+y$$ $$y'=\frac{2x+y}{2y-x}$$ - The slope of the tangent line at $(2,3)$ is $$y'=\frac{2\times2+3}{2\times3-2}=\frac{7}{4}$$ - The tangent line to the given curve at $(2,3)$ is $$y-3=\frac{7}{4}(x-2)$$ $$y-3=\frac{7}{4}x-\frac{7}{2}$$ $$y=\frac{7}{4}x-\frac{1}{2}$$ 3) Find the normal line at $(2,3)$: The normal line at $(2,3)$ would be perpendicular with the tangent line at $(2,3)$, so the product of the slopes of these two lines equal $-1$. So if we call the slope of the normal line at $(2,3)$ $k$, we would have $$k\times\frac{7}{4}=-1$$ $$k=-\frac{4}{7}$$ - The normal line to the given curve at $(2,3)$ therefore is $$y-3=-\frac{4}{7}(x-2)$$ $$y-3=-\frac{4}{7}x+\frac{8}{7}$$ $$y=-\frac{4}{7}x+\frac{29}{7}$$