# calculus

posted by .

x^2-x^2y=y^2-1

find all points on the curve where x=2. Show there is a horizontal tangent to the curve at one of the points.

• calculus -

sub x = 2 into your equation to get

4 - 4y = y^2 - 1
0 = y^2 + 4y - 5
(y+5)(y-1) = 0

so when x=2, y =-5 or when x=2, y = 1

so the points are (2,-5) and (2,1)

check them by subbing them back into the original equation, they work

• calculus -

I found the derivative of your equation implicitly and got
dy/dx = (2x+2xy)/(2y = x^2)

for a horizontal tangent this has to be zero, which means the numerator should be zero

subbing in (2,-5) makes the top zero, so there is a horizontal tangent at the point (2,_5)

## Similar Questions

1. ### calculus

dy/dx= (2x-2xy)/(x^2+2y) Find all points on the curve where x=2. Show there is a horizontal tangent to the curve at one of those points.
2. ### calculus

dy/dx= (2x-2xy)/(x^2+2y) Find all points on the curve where x=2. Show there is a horizontal tangent to the curve at one of those points.
3. ### calculus

dy/dx= (2x-2xy)/(x^2+2y) find all points on the curve where x=2. show there is a horizontal tangent to the curve at one of those points.
4. ### Calculus

a)The curve with equation: 2y^3 + y^2 - y^5 = x^4 - 2x^3 + x^2 has been linked to a bouncing wagon. Use a computer algebra system to graph this curve and discover why. b)At how many points does this curve have horizontal tangent lines?
5. ### Calculus

Consider the curve y^2+xy+x^2=15. What is dy/dx?
6. ### Calculus

Please help this is due tomorrow and I don't know how to Ive missed a lot of school sick Consider the curve given by the equation x^3+3xy^2+y^3=1 a.Find dy/dx b. Write an equation for the tangent line to the curve when x = 0. c. Write …
7. ### calculus

Given the curve defined by the equation y=cos^2(x) + sqrt(2)* sin(x) with domain (0,pi) , find all points on the curve where the tangent line to the curve is horizontal
8. ### Calculus

Consider the curve given by y^2 = 2+xy (a) show that dy/dx= y/(2y-x) (b) Find all points (x,y) on the curve where the line tangent to the curve has slope 1/2. (c) Show that there are now points (x,y) on the curve where the line tangent …
9. ### Calculus

Can someone show how this question is solved. Consider the curve given by the equation 2y^3 + y^2 - y^5 = x^4 -2x^3 +x^2 Find all points at which the tangent line to the curve is horizontal or vertical. Thanks!