Question

58–59. Tangent lines Find an equation of the line tangent to the following curves at the given point. Check your work with a graphing utility.x²/16 - y²/9 = 1; (20/3, -4)

Accepted Answer

Identify the given curve: $$\frac{x^2$}{16} - \frac{y^2$}{9} = 1$$ and the point of tangency $$\left(\frac{20}{3}, -4\right). $$Differentiate both sides of the equation implicitly with respect to $$x. $$Remember to apply the chain rule when differentiating terms involving $$y$$, treating $$y as a $$function of $$x. $$The differentiation will look like this: $$\frac{d}{dx}\left(\frac{x^2$}{16}\right) - \frac{d}{dx}\left(\frac{y^2$}{9}\right) = \frac{d}{dx}(1). $$Calculate each derivative: $$\frac{d}{dx}\left(\frac{x^2$}{16}\right) = \frac{2x}{16} = \frac{x}{8}$$, and for $$\frac{d}{dx}\left(\frac{y^2$}{9}\right)$$ use the chain rule: $$\frac{2y}{9} \cdot \frac{dy}{dx}. $$Set up the equation: $$\frac{x}{8} - \frac{2y}{9} \cdot \frac{dy}{dx} = 0. $$Solve for $$\frac{dy}{dx} ($$the slope of the tangent line) by isolating it: $$\frac{dy}{dx} = \frac{\frac{x}{8}}{\frac{2y}{9}} = \frac{x}{8} \cdot \frac{9}{2y} = \frac{9x}{16y}. $$Substitute the coordinates of the given point $$\left(\frac{20}{3}, -4\right)$$ into the expression for $$\frac{dy}{dx} to $$find the slope at that point. Then use the point-slope form of a line: $$y - y_1$ = m(x - x_1$)$$, where $$m is $$the slope and $$(x_1$, y_1$) is $$the point, to write the equation of the tangent line.

58–59. Tangent lines Find an equation of the line tangent to the following curves at the given point. Check your work with a graphing utility.
x²/16 - y²/9 = 1; (20/3, -4)

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