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Chapter 11: Problem 71

Find an equation of the line tangent to the following curves at the given point. $$y^{2}-\frac{x^{2}}{64}=1 ;\left(6,-\frac{5}{4}\right)$$

Short Answer

Expert verified
The equation of the line tangent to the curve at the specified point is \(y=\frac{-3}{10}(x-6)-\frac{5}{4}\).

Step by step solution

01

Write down the given curve equation and point

We are given the curve equation: $$y^{2}-\frac{x^{2}}{64}=1$$ And the point at which we need to find the tangent line is: $$P\left(6,-\frac{5}{4}\right)$$
02

Differentiate the equation implicitly with respect to x

To find the derivative of the given function, we will perform implicit differentiation: $$\frac{d}{dx} \left(y^{2}-\frac{x^{2}}{64}\right) = \frac{d}{dx}(1)$$ Differentiating both sides: $$2y\frac{dy}{dx}-\frac{2x}{64}\frac{dx}{dx}=0$$ Now, simplifying and solving for \(\frac{dy}{dx}\): $$\frac{dy}{dx}=\frac{x}{32y}$$
03

Find the slope of the tangent line at the given point

To find the slope of the tangent line at the given point, we'll plug the coordinates of the point into the derivative: $$\frac{dy}{dx}\Bigg|_{\left(6,-\frac{5}{4}\right)} = \frac{6}{32\left(-\frac{5}{4}\right)}$$ Simplifying this expression gives us: $$\frac{dy}{dx}\Bigg|_{\left(6,-\frac{5}{4}\right)} = -\frac{3}{10}$$ So the slope of the tangent line at the given point is \(-\frac{3}{10}\).
04

Find the equation of the tangent line using the point-slope form

The point-slope form of a linear equation is given by: $$y-y_1=m(x-x_1)$$ Where \((x_1,y_1)\) is the given point and \(m\) is the slope. Plugging the coordinates of the point and the slope into the equation: $$y-\left(-\frac{5}{4}\right)=-\frac{3}{10}(x-6)$$ Simplifying the equation to put it in slope-intercept form: $$y=\frac{-3}{10}(x-6)-\frac{5}{4}$$ This is the equation of the line tangent to the given curve at the specified point.

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