Question

Use implicit differentiation to find an equation of the tangent line to the cardioid at the point (0, -0.5). x^2 + y^2 = (2x^2 + 2y^2 - x)^2

Answer 1

Ok so where are you stuck at?
Were you able to find a y'?

Answer 2

This is one of those problems where you'll want to NOT try to solve for y' immediately.
After you take your derivative it's much easier to plug in your coordinate since one of them contains a zero.

Any trouble taking derivative of either side?

Answer 3

nope i get the derivative fine and when i plug in i get -1/3

Answer 4

Hmm I got y' = -1.
Lemme check my work again really quick.

Answer 5

Oh the left side of the equation, plugging in x=0 y=-1/2 gives you,\[\Large\bf\sf 2(0)+2\left(-\frac{1}{2}\right)y'\]Did you forget the negative on your y' ?

Answer 6

\[
2 y(x) y'(x)+2 x=2 \left(2
x^2+2 y(x)^2-x\right)
\left(4 y(x) y'(x)+4
x-1\right)\\
y'(x)=-\frac{2 \left(4 x^3-3
x^2+4 x
y(x)^2-y(x)^2\right)}{y(x)
\left(8 x^2+8 y(x)^2-4
x-1\right)}
\]

Answer 7

at x=0 and y[x]=-1/2
we get y'=-1

Answer 8

\[
y+\frac 1 2= -(x-0)\\
y=-x -\frac 1 2
\]
is the equation of the tangent.

Answer 9

yep confirmed... and graphed it to make sure.