Answer to Question #262149 in Calculus for onurky

Question

Use implicit differentiation to find the points where the parabola defined by

x^2−2xy+y^2−4y+16=0

has horizontal and vertical tangent lines.

The parabola has horizontal tangent lines at the point(s)

.

The parabola has vertical tangent lines at the point(s)

.

Answer 1

$x^2−2xy+y^2−4y+16=0........(i)$

take differentiation

$2udu-2udy-2ydu+2ydy-4dy=0\\(2u-2y)du+(2y-2u-4)dy=0\\(2u-2y)du=-(2y-2u-4)dy\\(2u-2y)du=(2u-2y+4)dy\\$

horizontal lines $\frac{dy}{du}=\frac{2x-2y}{2x-2y+4}=\frac{x-y}{x-y+2}$

$\frac{dy}{dx}=0\\\frac{x-y}{x-y+2}\\x=y$

for vertical tangent =$\frac{dy}{dx}$ =underdefined

or

$x-y+2=0\\y=x+2$

now solve the points on the parabola

substitute

x=y

in equation 1

$x^2-2x+x^2-4x+16=0\\-4x+16=0\\x=\frac{16}{4}\\x=4\\y=4$

horizontal(4,4)

for vertical tangent lines y=x+2

put in equation 1

$x^2-2x(x+2)+(x+2)^2-4(x+2)+16=0\\-4x=12=0\\-4x=-12\\x=3\\y=3+2=5$

vertical tangent lines(3,5)

the parabola has horizontal tangent lines at (4,4)

the parabola has vertical tangent lines at (3,5)