Question

Help Please!(:
Find the equation to the tangent line to the curve
2(x^2+y^2)^2=25(x^2-y^2)
at the point (-3,1)

Answer 1

@tkhunny

Answer 2

Rule #1 - Make sure the point IS ON the curve.

Find the Implicit Derivative. What's preventing you from doing it?

Answer 3

it's not in y-intercept form?

Answer 4

What does that even mean? I'll do the right-hand side.

\(Given\; y = f(x),\;\dfrac{d}{dx}\left[25(x^{2}-y^{2})\right] = 25\left(2x - 2y\dfrac{dy}{dx}\right)\)

Okat, you do the left-hand side.

Answer 5

\[2(x^2+y^2)^2 = 4(x^2+y^2)(2x+2y)\]

Answer 6

@tkhunny

Answer 7

So close!! One last little thing next to your 2y on the end. Take a look at mine.

Answer 8

dy/dx?

Answer 9

That's it. Write all that in one ugly equation and solve for dy/dx.

Answer 10

how do I solve fr dy/dx? I'm confused

Answer 11

Please, just write the two expressions we have created. Each of them has a dy/dx element. Solve for it just like you have been solving for things since Algebra I.

Answer 12

\[4(x^2+y^2)(dy/dx)=25(2x-27)(dy/dx)\]

Answer 13

idk what to do after that.

Answer 14

sorry theres a lot of typos. lemme fix that

Answer 15

\[4(x^2+y^2)(2x+2y)(dy/dx)=25(2x-2y)(dy/dx)\]

Answer 16

Close. Fix the rest and you'll be on your way.

\(4(x^{2} + y^{2})(2x + 2y(dy/dx)) = 25(2x - 2y(dy/dx))\)

Please use your best algebra to solve for dy/dx.

Answer 17

okay, i get that i need to solve for dy/dx, but how? do I add values together? i don't understand what solving for dy/dx means in this term.

Answer 18

Algebra? Isolate it, using whatever legal operations are required. It may help to substitute for it on your first try. Write the expression again, only this time, simply write 'z' instead of dy/dx. Then, solve for z.

Answer 19

Did anyone else get
y = (9/13) x + 40/13 ?