What is the tangent of (x^2+y^2)^2=4x^2y,(1,1) ?

The tangent of (x^2+y^2)^2=4x^2y,(1,1) is y=1

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tangent of (x^2+y^2)^2=4x^2y,(1,1)

tangent of

(x^{2} + y^{2})^{2} = 4 x^{2} y, (1, 1)

y = 1

Solution steps

Find the slope of

(x^{2} + y^{2})^{2} = 4 x^{2} y : \frac{d y}{d x} = \frac{x ( - y ^{2} + 2 y - x ^{2} )}{- x ^{2} + y ^{3} + x ^{2} y}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 0

Find the line with slope m=

0

and passing through

(1, 1) : y = 1

y = 1

\int_{0}^{2} \frac{2 x}{( x ^{2} + 2 ) ^{3}} d x

\int e^{3 x} - 18 e^{- 4 x} d x

\frac{d}{d x} (\frac{5 sin ( x )}{1 + cos ( x )})

\int_{0}^{3} (∣ 2 x - 3 ∣) d x

x \to + 4 lim (6)

What is the tangent of (x^2+y^2)^2=4x^2y,(1,1) ?
The tangent of (x^2+y^2)^2=4x^2y,(1,1) is y=1