What is the normal of 5x^2+9y^2=36,(6,4) ?

The normal of 5x^2+9y^2=36,(6,4) is Normal line to 5x^2+9y^2=36, with a tangent of point (6,4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

normal of 5x^2+9y^2=36,(6,4)

normal of

5 x^{2} + 9 y^{2} = 36, (6, 4)

Normal line to 5 x^{2} + 9 y^{2} = 36, with a tangent point (6, 4)

Solution steps

5 x^{2} + 9 y^{2} = 36 : y = \frac{36 - 5 x ^{2}}{3}, y = - \frac{36 - 5 x ^{2}}{3}

Slope of the normal line at

(6, 4) : m = \frac{- 36 - 5 x ^{2} - 12}{3 ( x - 6 )}

Find the slope of

y = \frac{36 - 5 x ^{2}}{3}, y = - \frac{36 - 5 x ^{2}}{3} : m_{y} = \frac{5 x}{3 - 5 x ^{2} + 36}

Compute the slope of the normal line:

m = - \frac{3 - 5 x ^{2} + 36}{5 x}

What is the normal of 5x^2+9y^2=36,(6,4) ?
The normal of 5x^2+9y^2=36,(6,4) is Normal line to 5x^2+9y^2=36, with a tangent of point (6,4)