Question

Find an equation of the line tangent to the following curves at the given point. $$x^{2}=-6 y ;(-6,-6)$$

Short answer

Question: Find the equation of the tangent line to the curve \(x^2 = -6y\) at the point \((-6, -6)\). Answer: The equation of the tangent line to the curve at the point \((-6, -6)\) is \(y = 2x + 6\).

Step-by-step solution

Find the derivative of the function with respect to x

First, we need to differentiate the given equation \(x^2 = -6y\) with respect to \(x\). We can do this using implicit differentiation: $$\frac{d}{dx}(x^2) = \frac{d}{dx}(-6y)$$ $$2x = -6\frac{dy}{dx}$$ Now, solve for \(\frac{dy}{dx}\): $$\frac{dy}{dx} = -\frac{1}{3}x$$

Evaluate the derivative at the given point

We have the derivative of the function with respect to \(x\): \(\frac{dy}{dx} = -\frac{1}{3}x\). Now, we need to evaluate the derivative at the point \((-6, -6)\) to find the slope of the tangent line at this point. $$m = \frac{dy}{dx}\Big|_{(-6,-6)} = -\frac{1}{3}(-6)$$ $$m = 2$$ So, the slope of the tangent line at the point \((-6, -6)\) is 2.

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

Now that we have the slope (\(m=2\)) of the tangent line and the point \((-6, -6)\), we can use the point-slope form of a linear equation to find the equation of the tangent line: $$y - y_1 = m(x - x_1)$$ Substitute the given point and slope into the equation: $$y - (-6) = 2(x - (-6))$$ Simplify the equation: $$y + 6 = 2(x + 6)$$ Distribute the 2: $$y + 6 = 2x + 12$$ $$y = 2x + 6$$ So, the equation of the tangent line to the curve \(x^2 = -6y\) at the point \((-6, -6)\) is \(y = 2x + 6\).