Question

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

Question: Find the equation of the line tangent to the curve \(y^2 - \frac{x^2}{64} = 1\) at the point \(\left(6, -\frac{5}{4}\right)\). Answer: The equation of the tangent line to the curve at the given point is: \(y = -\frac{3}{20}(x - 6) - \frac{5}{4}\).

Step-by-step solution

Differentiate the curve implicitly with respect to x

Start by differentiating the given equation: $$y^{2}-\frac{x^{2}}{64}=1$$ Differentiate both sides with respect to x: $$\frac{d}{dx}(y^2) - \frac{1}{64}\frac{d}{dx}(x^2) = \frac{d}{dx}(1)$$ Apply the power rule and chain rule: $$2y\frac{dy}{dx} - \frac{1}{32}x = 0$$

Solve for dy/dx

Rearrange the equation from Step 1 to solve for dy/dx, which will give us the slope of the tangent line: $$2y\frac{dy}{dx} = \frac{1}{32}x$$ $$\frac{dy}{dx} = \frac{x}{64y}$$

Evaluate dy/dx at the given point

Plug the given point \(\left(6,-\frac{5}{4}\right)\) into the equation we found in Step 2 and evaluate the derivative (slope) at this point. $$\frac{dy}{dx} = \frac{6}{64\left(-\frac{5}{4}\right)}$$ Simplify the expression: $$\frac{dy}{dx} = -\frac{3}{20}$$

Use point-slope form to find the equation of the tangent line

Use the point-slope form of the equation of a line, which is given by: $$y - y_1 = m(x - x_1)$$ Where \((x_1, y_1)\) is our given point \(\left(6,-\frac{5}{4}\right)\) and \(m\) is the slope we found in Step 3 (\(-\frac{3}{20}\)): $$y + \frac{5}{4} = -\frac{3}{20}(x - 6)$$

Write the equation in slope-intercept form, if desired

Rearrange the equation from Step 4 to put it in slope-intercept form (\(y = mx + b\)) by distributing the slope and isolating y: $$y = -\frac{3}{20}(x - 6) - \frac{5}{4}$$ And that's the equation of the line tangent to the given curve at the given point!