what is the equation of the tangent line to the curve (a lemniscate) 2(x^2+y^2)^2=25(x^2-y^2) at point (3,1)?

Question

the equation of this tangent line can be written in the form y=mx+b

hint: you may use that: (x^2+y^2)^2=x^4+y^4+2x^2y^2 before differentiating.

Answer 1

2(x^2+y^2)^2=25(x^2-y^2)

4(x^2+y^2)(2x+2yy')=25(2x-2yy')
x=3
y=1
4(9+1)(6+2y')=25(6-2y')
8(6+2y')=5(6-2y')
48+16y'-30+10y'=0
26y'=-18
y'=-9/13
y-1=-9/13(x-3)

y=-9/13(x-3)+1

Answer 2

what's the curve equation to stumble on the tangent line at a factor (25,5) on it? EDIT: [As consistent with the added info, which i ought to relook purely after approximately 9 hours of formerly presentation] a million) Differentiating the given one, dy/dx = a million/(2?x) 2) At x = 25, dy/dx = a million/(2?25) = a million/10; that's the slope of the tangent line on the given factor; that's via the geometrical definition of differentiation] 3) making use of Slope-factor sort, the equation of the tangent line at (25,5) is: y - 5 = (a million/10)(x - 25) increasing and simplifying, the equation is: x - 10y + 25 = 0

Answer 3

Lol you are probably in the same class as me cuz i have the same question ... well this is how u do it ....

the answer is y=-9x/13+40/13

solution
2(x^2+y^2)^2=25(x^2-y^2)
4(x^2+y^2)(2x+2yy')=25(2x-2yy')
Sunstitute x=3, y=1
40(6+2y')=25(6-2y')
130y'=-90
y'=-9/13
Equation of tangent line is
y-1=-9/13(x-3)

Answer 4

yeah um i suck at this stuff (Maths) i am bloody gonna drop out of maths in school oneday!

Answer 5

what isn't it?