How do you find the equations for the normal line to $x^{2} + y^{2} = 20$ $x^2+y^2=20$ through (2,4)?

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How do you find the equations for the normal line to $x^{2} + y^{2} = 20$ $x^2+y^2=20$ through (2,4)?

2 Answers

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Answer 1 · 2017-04-23T03:49:42

$y = 2 x$ $y=2x$

Explanation:

$y - 0 = \frac{4 - 0}{2 - 0} (x - 0)$ $y-0 = (4-0)/(2-0)(x-0)$

$y = 2 x$ $y=2x$

:)))

Answer 2 · 2017-04-23T03:55:38

The equation is $y = 2 x$ $y =2x$ .

Explanation:

The derivative of this relation is given by

$2 x + 2 y (\frac{d y}{d x}) = 0$ $2x + 2y(dy/dx) = 0$

$2 y (\frac{d y}{d x}) = - 2 x$ $2y(dy/dx) = -2x$

$\frac{d y}{d x} = - \frac{x}{y}$ $dy/dx= -x/y$

The slope of the tangent line is

$m_{tangent} = - \frac{2}{4} = - \frac{1}{2}$ $m_"tangent" = -2/4 = -1/2$

Therefore, the slope of the normal line is $2$ $2$ .

$y - 4 = 2 (x - 2)$ $y - 4 = 2(x - 2)$

$y - 4 = 2 x - 4$ $y- 4= 2x - 4$

$y = 2 x$ $y = 2x$ , as obtained in the other answer

Hopefully this helps!

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How do you find the equations for the normal line to $x^{2} + y^{2} = 20$ $x^2+y^2=20$ through (2,4)?

2 Answers

Explanation:

Explanation:

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How do you find the equations for the normal line to $x^{2} + y^{2} = 20$ $x^2+y^2=20$ through (2,4)?