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)

Question

anonymous · Answer

Are you okay with implicit differentiation?

dessyj1 · Answer

yes i am

anonymous · Answer

So did you find dy/dx?

dessyj1 · Answer

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

dessyj1 · Answer

when i plugged in the values.

dessyj1 · Answer

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

anonymous · Answer

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

dessyj1 · Answer

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

anonymous · Answer

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

dessyj1 · Answer

how would we find the fixed point on the curve?

anonymous · Answer

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.

anonymous · Answer

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

dessyj1 · Answer

thank you