Question

1. (5 pts.) consider the curve described by the equation 2y² - x² = 4y. verify that the point (4, -2) lies on this curve, and find y at that point.

Explanation:

Step1: Verify the point lies on the curve

Substitute \(x = 4\) and \(y=-2\) into the equation \(2y^{2}-x^{2}=4y\).
Left - hand side: \(2(-2)^{2}-4^{2}=2\times4 - 16=8 - 16=-8\).
Right - hand side: \(4\times(-2)=-8\). Since the left - hand side equals the right - hand side, the point \((4,-2)\) lies on the curve.

Step2: Differentiate the equation implicitly

Differentiate \(2y^{2}-x^{2}=4y\) with respect to \(x\).
Using the chain rule and power rule, we have \(4y\frac{dy}{dx}-2x = 4\frac{dy}{dx}\).

Step3: Solve for \(\frac{dy}{dx}\)

Rearrange the terms to isolate \(\frac{dy}{dx}\):
\(4y\frac{dy}{dx}-4\frac{dy}{dx}=2x\).
Factor out \(\frac{dy}{dx}\): \(\frac{dy}{dx}(4y - 4)=2x\).
So, \(\frac{dy}{dx}=\frac{2x}{4y - 4}=\frac{x}{2y - 2}\).

Step4: Evaluate \(\frac{dy}{dx}\) at the point \((4,-2)\)

Substitute \(x = 4\) and \(y=-2\) into \(\frac{dy}{dx}\):
\(\frac{dy}{dx}=\frac{4}{2\times(-2)-2}=\frac{4}{-4 - 2}=-\frac{2}{3}\).

Answer:

The point \((4,-2)\) lies on the curve and \(\frac{dy}{dx}=-\frac{2}{3}\) at that point.