## University Calculus: Early Transcendentals (3rd Edition)

1) The point $(\sqrt3,2)$ is on the curve. 2) The tangent line to the curve at $(\sqrt3,2)$ is $y=2$. 3) The normal line to the curve at $(\sqrt3,2)$ is $x=\sqrt3$.
$$x^2-\sqrt3xy+2y^2=5$$ 1) Verify that the point $(\sqrt3,2)$ is on the curve We substitute the coordinates of the point $(\sqrt3,2)$ into the formula of the function to see if it equals $5$ or not: $$(\sqrt3)^2-\sqrt3\times\sqrt3\times2+2\times2^2=3-6+8=5$$ Since it equals $5$, the point is on the curve. 2) Find the tangent at $(\sqrt3,2)$: - Find the derivative of the function with implicit differentiation: $$2x-\sqrt3(y+xy')+4yy'=0$$ $$2x-\sqrt3y-\sqrt3xy'+4yy'=0$$ $$y'(4y-\sqrt3x)=\sqrt3y-2x$$ $$y'=\frac{\sqrt3y-2x}{4y-\sqrt3x}$$ - The slope of the tangent line at $(\sqrt3,2)$ is $$y'=\frac{\sqrt3\times2-2\sqrt3}{4\times2-\sqrt3\times\sqrt3}=\frac{0}{8-3}=0$$ - The tangent line to the given curve at $(\sqrt3,2)$ is $$y-2=0(x-\sqrt3)$$ $$y-2=0$$ $$y=2$$ 3) Find the normal line at $(-1,0)$: Since the tangent line at $(\sqrt3,2)$ is the horizontal line $y=2$, the normal line at $(\sqrt3,2)$, being perpendicular with the tangent line, would be a vertical line with the form $x=x_0$ And as the normal line crosses the point $(\sqrt3,2)$, $x_0=\sqrt3$, meaning the normal line in need to find is $$x=\sqrt3$$