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

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

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

tangent of

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

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

Solution steps

Find the slope of

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

Find the line with slope m=

- \frac{1}{2}

and passing through

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

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

\int (5 - 4 e^{x} + \frac{3}{x ^{2}}) d x

h \to 0 lim (\frac{( 2 h ^{2} + 6 h )}{h})

x \to 0 lim (f (x) \cdot g (x))

\frac{d}{d x} (\frac{- x ^{2}}{4})

f (x) = (4 z^{4} + 2 z^{2} + 1) (2 z^{3} - z)

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