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

The normal of x^2-xy+y^2=3,(1,2) is x=1

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normal of x^2-xy+y^2=3,(1,2)

normal of

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

x = 1

Solution steps

Find the slope of

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

Compute the slope of

x^{2} - x y + y^{2} = 3

(1, 2) : m = 0

Compute the slope of the perpendicular line:

m_{p} = \infty

Find the line with slope m=

\infty

and passing through

(1, 2) : x = 1

x = 1

lo g_{2} (x lo g_{5} (x))

y = \frac{2}{sin ( x ) + cos ( x )}, (0, 2)

n = 1 \sum \infty n (0.9)^{n}

\frac{d}{d x} (x \cdot sin^{2} (x))

\frac{d}{d x} (\frac{b}{( 1 + e ^{(- i x + r)} )} + z)

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