## dessyj1 one year ago Given the curve x^2+3xy-2y^2=2 Find the equation of the line tangent to the curve at the point (1,1) then Find the co-ordinates of all other points on this curve with the slope equal to the slope at (1,1)

1. anonymous

Are you okay with implicit differentiation?

2. dessyj1

yes i am

3. anonymous

So did you find dy/dx?

4. dessyj1

i isolated to y-prime and got the slope to be 5

5. dessyj1

when i plugged in the values.

6. dessyj1

I just do not know what to do with the follow up question.

7. anonymous

Okay, so I assume you got the tangent line as well then?

8. dessyj1

Yes, the equation of the tangent was 0=5x-4-y

9. anonymous

Alrighty. So let's see $\frac{ dy }{ dx } = \frac{ 2x+3y }{ 4y-3x }$ And we need all points which also give us a slope of 5. $5 = \frac{ 2x+3y }{ 4y-3x }$ 20y - 15x = 2x + 3y 17y = 17x y = x So we can get a slope of 5, whenever y = x

10. dessyj1

how would we find the fixed point on the curve?

11. anonymous

Right, we would need to see which of those points actually exist on the curve. Well, we know x must equal to y to get the proper slope. So replace y with x in the original equation.

12. anonymous

You'd get x^2 + 3x^2 - 2x^2 = 2 2x^2 = 2 x = 1 or -1

13. dessyj1

thank you