Question

Find the equation of the tangent line to the curve 2(x^2+y^2)^2 = 25(x^2-y^2) at the point ( 3 , -1 ). The equation of this tangent line can be written in the form y = mx+b. Find m and b.

Answer 1

I still have no idea what to do :\

Answer 2

Tke Equation solving technique in use

Answer 3

I think m=85/37 and b= -292/37

Answer 4

@FatihSava how did you get that answer? Mind explaining? Thanks a lot.

Answer 5

sadly, but it's true. thanks for your medal

Answer 6

Hey man its the best answer i've gotten in the past hour so you deserved it haha.

Answer 7

ok, glad to hear it. I am in bad mood, cannot think anything to help. sorry for that

Answer 8

@Mettalknight do you know implicit derivative?

Answer 9

@Hoa I sort of understand how it works but not completely. :\

Answer 10

ok, since you give me medal, no way to say no to you, right? we do together but I don't promise that we get the right one.
to me, take implicit derivative, and plug (3,-1) in x and y at that derivative to get m
b is y intercept. that mean x =0, plug that value into the original function to get y . that y is b. try it

Answer 11

So I tried to derive 2(x^2+y^2)^2 = 25(x^2-y^2) and got (-4x^3 - 4xy^2 + 25x) / (y (4x^2 + y^2 + 25)) and then I plugged in (3,-1) and got 1.12227 and that's incorrect :\

Answer 12

I used derivative of implicit functions.

Answer 13

You have a point, (3, -1) given. Now all you need is the slope of the tangent, which is the derivative dy/dx at x = 3. Using implicit differentiation gives us:\[2(x^2+y^2)^2=25(x^2-y^2) \rightarrow 2[2(x^2+y^2)(2x+2y \frac{ dy }{ dx })]=25(2x-2y \frac{ dy }{ dx })\]\[ 8(x^2+y^2)(x+y \frac{ dy }{ dx })=50(x-y \frac{ dy }{ dx })\]\[8x^3+8x^2\frac{ dy }{ dx }+8xy^2+8y^3\frac{ dy }{ dx }=50x-50y\frac{ dy }{ dx }\]\[8x^3+8xy^2-50x=-50y\frac{ dy }{ dx }-8x^2\frac{ dy }{ dx }-8y^3\frac{ dy }{ dx }\]\[8x^3+8xy^2-50x=\frac{ dy }{ dx }(-50y-8x^2-8y^3)\]\[\frac{ dy }{ dx } = \frac{ 8x^3+8xy^2-50x }{ -50y-8x^2-8y^3 }\]Now let's take the derivative by plugging in x = 3.\[f'(3) = \frac{ 8(3)^3+8(3)(-1)^2-50(3) }{ -50(-1)-8(3)^2-8(-1)^3 }=\frac{ 216+24-150 }{ 50-72+8 }=\frac{ 90 }{ -14 }=-\frac{45 }{ 7 }\]So the slope of the tangent is -45/7 at (3, -1). We use this slope and the given point to find the equation of the tangent.\[y=mx+b \rightarrow -1=-\frac{ 45 }{ 7 }(3)+b \rightarrow b = \frac{ 128 }{ 7 }\]Now that we have solved for b, we plug that in and write the final equation of the tangent line:\[y=-\frac{ 45 }{ 7 }x+\frac{ 128 }{ 7 }\]
@Mettalknight

Answer 14

Sorry, I forgot the y in "-8x^2" in the denominator of dy/dx. It should be "-8x^2y" in the denominator of dy/dx. Just plug that extra "y" in there and solve again for the slope and get the equation of the tangent once again the way I did. Sorry for that tiny and stupid mistake -.-.

@Mettalknight

Answer 15

@genius12 Thank you so much! You're the first person to actually answer it right. Now I just need to make sure I understand what you did.

Answer 16

Thanks and no problem :D fan me? lol @Mettalknight

Answer 17

@genius12 Done. Thank you again so much for your time and help :)

Answer 18

Yup. Btw, you did read my post on that extra y I forgot to add right? lol @Mettalknight
So you will need to reevaluate the slope after adding the extra y and then will have to find the equation for the tangent once again. Stupid mistake =[ @Mettalknight

Answer 19

@genius12 yes I did don't worry about it. The final slope is 90/130 now I just need to find b

Answer 20

Yes exactly.

Answer 21

Simplifying gives 9/13 @Mettalknight