Find the equation of the line tangent to the curve 2(x^2+y^2)^2=25(x^2+y^2) at a point (3,-1) written in the form y = mx+b
ahhhh, just finished all that stuff below, but
it appears that the given point is NOT on the curve, so the question is bogus.
Which shows that the first thing you should check for this kind of question is to check if the given point lies on the curve.
2(x^2+y^2)^2=25(x^2+y^2)
2(x^2+y^2)^2=25x^2 + 25y^2
4(x^2 + y^2) (2x + 2y dy/dx) = 50x + 50y dy/dx
so for the given point (3,-1)
4(9 + 1)(6 - 2 dy/dx) = 150 - 50 dy/dx
40(6 - 2 dy/dx) = 150 - 50 dy/dx
240 - 80 dy/dx = 150 - 50 dy/dx
-30 dy/dx = -90
dy/dx = 3
equation:
y + 1 = 3(x - 3)
or
y = 3x - 10
or, I should have seen that ....
divide both sides by x^2 + y^2
2(x^2 + y^2) = 25
x^2 + y^2 = 25/2
2x + 2y dy/dx = 0
dy/dx = -2x/2y = -x/y
for the given point , dy/dx = -6/-2 = 3
continue as before.
or
after reducing, we have the circle x^2 + y^2 = 25.
It is here that I noticed that the given point is not even on the curve!!!