What is the tangent of 3x^2+y^2=1,(1/2 ,(-1)/2) ?

The tangent of 3x^2+y^2=1,(1/2 ,(-1)/2) is y=3x-2

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tangent of 3x^2+y^2=1,(1/2 ,(-1)/2)

tangent of

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

y = 3 x - 2

Solution steps

Find the slope of

3 x^{2} + y^{2} = 1 : \frac{d y}{d x} = - \frac{3 x}{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 = 3

Find the line with slope m=

3

and passing through

(\frac{1}{2}, \frac{- 1}{2}) : y = 3 x - 2

y = 3 x - 2

y^{^{''}} - y^{^{'}} + 12 y = 0

\frac{d}{d x} (\frac{( e ^{x} )}{1 - e ^{- x}})

\int - \frac{x}{x ^{2} + y ^{2}} d x

y^{^{'}} + \frac{y}{t} = 2 e^{t}

x^{2} + 6 x

What is the tangent of 3x^2+y^2=1,(1/2 ,(-1)/2) ?
The tangent of 3x^2+y^2=1,(1/2 ,(-1)/2) is y=3x-2