Posted by Lee on Sunday, October 30, 2011 at 8:40pm.
just plug and chug:
4(x^2 + y^2)(2x + 2yy') = 9(2x - 2yy')
Now, the tangent is horizontal when y' = 0
4(x^2 + y^2)(2x) = 9(2x)
8x^3 + 8xy^2 = 18x
2x(4x^2 + 4y^2 - 9) = 0
So, either x=0
substitute back into original equation:
2y^4 = -9y^2
y=0
or,
4x^2 + 4y^2 = 9
x^2 = (9 - 4y^2)/4
Substitute that back into the original equation
2((9 - 4y^2)/4 + y^2)^2 = 9((9 - 4y^2)/4 - y^2)
Expand the binomial and solve for y^2
Tan^3(xy^2+y)=x
Use derivative
Find .... Tan^3(xy^2+y)=x
Use derivative
Related Questions
Calculus - Find the equation of the tangent line to the curve (a lemniscate) 2(x...
Calculus. I need help! - HARDER PARTS WAS 3(x^2+y^2)^2=26(y^2+y^2) Find the ...
Calculus - Verify? - Find the slope of the tangent line to the curve (a ...
calculus - the lemniscate revolves about a tangent at the pole find sufaces of ...
Calculus - Please help this is due tomorrow and I dont know how to Ive missed a ...
Implicit Differentiation - Find the slope of the tangent line to the curve (a ...
Calculus - the curve: (x)(y^2)-(x^3)(y)=6 (dy/dx)=(3(x^2)y-(y^2))/(2xy-(x^3)) a...
Calculus - Consider the curve y^2+xy+x^2=15. What is dy/dx? Find the two points ...
calculus - Given the curve defined by the equation y=cos^2(x) + sqrt(2)* sin(x) ...
Calculus - a)The curve with equation: 2y^3 + y^2 - y^5 = x^4 - 2x^3 + x^2 has ...
For Further Reading