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Calculus / Calculus: Early Transcendentals 1 / Chapter 10.4 / Problem 77E

Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett

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Tangent lines for an ellipse Show that an equation of the

Chapter 9, Problem 77E

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QUESTION:

Show that an equation of the line tangent to the ellipse \(x^2/a^2+y^2/b^2=1\) at the point \((x_0,\ y_0)\) is

\(\frac{xx_0}{a^2}+\frac{yy_0}{b^2}=1\).

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QUESTION:

Show that an equation of the line tangent to the ellipse \(x^2/a^2+y^2/b^2=1\) at the point \((x_0,\ y_0)\) is

\(\frac{xx_0}{a^2}+\frac{yy_0}{b^2}=1\).

ANSWER:

Solution 77E
Step 1:

To find the tangent line at a point, first we need to find the slope( $\frac{dy}{dx}$ ) at the point.

Differentiating $\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1$ with respect to x to obtain $\frac{dy}{dx}$ , we get
$\frac{d}{dx}(\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}})=\frac{d}{dx}(1)$
$\frac{d}{dx}\frac{{x}^{2}}{{a}^{2}}+\frac{d}{dx}\frac{{y}^{2}}{{b}^{2}}=0$
$\frac{{2x}^{\ }}{{a}^{2}}+\frac{2y}{{b}^{2}}\frac{dy}{dx}=0$
$\frac{dy}{dx}=\frac{-{b}^{2}x}{{a}^{2}{y}^{\ }}$

Using the above result, the slope(m) at point $\left({{x}_{0},{y}_{0}}\right)$ is $m=\frac{-{b}^{2}{x}_{0}}{{a}^{2}{{y}_{0}}^{\ }}$

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