What is the normal of y=4x^2+3y^2+8x-6y-45,(-2,-3) ?

The normal of y=4x^2+3y^2+8x-6y-45,(-2,-3) is y=3.15261…x+3.30522…

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normal of y=4x^2+3y^2+8x-6y-45,(-2,-3)

normal of

y = 4 x^{2} + 3 y^{2} + 8 x - 6 y - 45, (- 2, - 3)

y = 3.15261 \dots x + 3.30522 \dots

Solution steps

y = 4 x^{2} + 3 y^{2} + 8 x - 6 y - 45 : y = \frac{7 + - 48 x ^{2} - 96 x + 589}{6}, y = \frac{7 - - 48 x ^{2} - 96 x + 589}{6}

Slope of the normal line at

(- 2, - 3) : m = \frac{- - 48 x ^{2} - 96 x + 589 + 25}{6 ( x + 2 )}

Find the slope of

y = \frac{7 + - 48 x ^{2} - 96 x + 589}{6}, y = \frac{7 - - 48 x ^{2} - 96 x + 589}{6} : m_{y} = \frac{8 ( x + 1 )}{- 48 x ^{2} - 96 x + 589}

Compute the slope of the normal line:

m = - \frac{- 48 x ^{2} - 96 x + 589}{8 ( x + 1 )}

Solve

\frac{- - 48 x ^{2} - 96 x + 589 + 25}{6 ( x + 2 )} = - \frac{- 48 x ^{2} - 96 x + 589}{8 ( x + 1 )} : x \approx - 1.96496 \dots

Find the normal line at

x = - 1.96496 \dots : y = 3.15261 \dots x + 3.30522 \dots

y = 3.15261 \dots x + 3.30522 \dots

What is the normal of y=4x^2+3y^2+8x-6y-45,(-2,-3) ?
The normal of y=4x^2+3y^2+8x-6y-45,(-2,-3) is y=3.15261…x+3.30522…