Question

1. consider the curve defined by ( x^{2}+x y + 2 y^{2}=32 )

(a) find the ( y ) intercepts of the curve defined above.
(b) write an expression for the slope of the curve at the point ( (x, y) ), that is
find ( \frac{d y}{d x} )
(c) determine whether the lines tangent to the curve at the ( y ) intercept of
the curve are parallel. show the analysis that leads to your conclusion.

Explanation:

Step1: Find \( y \)-intercepts

For \( y \)-intercepts, set \( x = 0 \) in the equation \( x^{2}+xy + 2y^{2}=32 \).
Substituting \( x = 0 \), we get \( 2y^{2}=32 \).
Divide both sides by 2: \( y^{2}=\frac{32}{2}=16 \).
Take square roots: \( y=\pm\sqrt{16}=\pm4 \).

Step2: Differentiate implicitly

Differentiate \( x^{2}+xy + 2y^{2}=32 \) with respect to \( x \).
Using the sum rule and product rule:

• The derivative of \( x^{2} \) is \( 2x \).
• For \( xy \), use the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \( u = x \), \( v = y \). So its derivative is \( y + x\frac{dy}{dx} \).
• For \( 2y^{2} \), use the chain rule. Its derivative is \( 4y\frac{dy}{dx} \).
• The derivative of 32 is 0.

So, \( 2x+y + x\frac{dy}{dx}+4y\frac{dy}{dx}=0 \).
Group terms with \(\frac{dy}{dx}\): \( (x + 4y)\frac{dy}{dx}=-2x - y \).
Solve for \(\frac{dy}{dx}\): \(\frac{dy}{dx}=\frac{-2x - y}{x + 4y}\).

Step3: Check slopes at \( y \)-intercepts

At \( (0,4) \):
Substitute \( x = 0 \), \( y = 4 \) into \(\frac{dy}{dx}=\frac{-2x - y}{x + 4y}\).
\(\frac{dy}{dx}=\frac{-2(0)-4}{0 + 4(4)}=\frac{-4}{16}=-\frac{1}{4}\).
At \( (0,-4) \):
Substitute \( x = 0 \), \( y=-4 \) into \(\frac{dy}{dx}=\frac{-2x - y}{x + 4y}\).
\(\frac{dy}{dx}=\frac{-2(0)-(-4)}{0+4(-4)}=\frac{4}{-16}=-\frac{1}{4}\).

Answer:

(a) The \( y \)-intercepts are \( (0,4) \) and \( (0, - 4) \).
(b) \(\frac{dy}{dx}=\frac{-2x - y}{x + 4y}\).
(c) The slopes of the tangent lines at \( (0,4) \) and \( (0,-4) \) are both \(-\frac{1}{4}\). So the tangent lines at the \( y \)-intercepts are parallel.