# Thread: Vertical Tangent to a Curve

1. ## Vertical Tangent to a Curve

I'm having trouble with this problem.

The tangent to the curve (y^2)-xy+9=0 is vertical when
(a) y= 0
(c) y= 1/2
(d) y= +/- 3
(e) none of these

I took the derivative of the equation and solved for x' and I got
x'= (x-2y)/(-y)
and then I set it to 0 but I get
x-2y=0
and I don't know what to do from there.

2. x - 2y = 0

x = 2y

substitute 2y for x in the original equation ... solve for y, and then determine x.

3. Hello, summermagic!

The tangent to the curve: $y^2-xy+9\:=\:0$ is vertical when:

. . $(a)\;y= 0 \qquad (b)\;y= \pm\sqrt{3} \qquad (c)\;y= \tfrac{1}{2} \qquad (d)\;y= \pm3 \qquad (e)\text{ none of these}$

Differentiate implicitly: . $2yy' - xy' - y \:=\:0\quad\Rightarrow\quad y' \:=\:\frac{y}{2y-x}$

The tangent is vertical when the denominator is zero:
. . $2y - x \:=\:0 \quad\Rightarrow\quad x \:=\:2y$

Substitute: . $y^2 - (2y)y + 9 \:=\:0 \quad\Rightarrow\quad -y^2 + 9 \:=\:0$

Therefore: . $y^2 \:=\:9 \quad\Rightarrow\quad y \:=\:\pm3$ . . . answer (d)

4. Okay, thank you!

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### points on curve where tangent is vertical y^2-xy 9

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