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)

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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)

Mathematics
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Are you okay with implicit differentiation?
yes i am
So did you find dy/dx?

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i isolated to y-prime and got the slope to be 5
when i plugged in the values.
I just do not know what to do with the follow up question.
Okay, so I assume you got the tangent line as well then?
Yes, the equation of the tangent was 0=5x-4-y
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
how would we find the fixed point on the curve?
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.
You'd get x^2 + 3x^2 - 2x^2 = 2 2x^2 = 2 x = 1 or -1
thank you

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