What is the tangent of x^2+2xy+4y^2=12,(2,1) ?

The tangent of x^2+2xy+4y^2=12,(2,1) is y=-1/2 x+2

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tangent of x^2+2xy+4y^2=12,(2,1)

tangent of

x^{2} + 2 x y + 4 y^{2} = 12, (2, 1)

y = - \frac{1}{2} x + 2

Solution steps

Find the slope of

x^{2} + 2 x y + 4 y^{2} = 12 : \frac{d y}{d x} = - \frac{x + y}{x + 4 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 = - \frac{1}{2}

Find the line with slope m=

- \frac{1}{2}

and passing through

(2, 1) : y = - \frac{1}{2} x + 2

y = - \frac{1}{2} x + 2

\int (- x^{2} + 6)^{2} - (x^{2} + 4 x)^{2} d x

2 x^{3}

- 27

\frac{1}{s ( s ^{2} + 16 )}

\frac{d}{d s} (\frac{1}{s ^{2} + 1})

What is the tangent of x^2+2xy+4y^2=12,(2,1) ?
The tangent of x^2+2xy+4y^2=12,(2,1) is y=-1/2 x+2