1. consider the curve defined by ( x^{2}+x y + 2 y^{2}=32 )\n(a) find the ( y ) intercepts of the curve…

1. consider the curve defined by ( x^{2}+x y + 2 y^{2}=32 )\n(a) find the ( y ) intercepts of the curve defined above.\n(b) write an expression for the slope of the curve at the point ( (x, y) ), that is\nfind ( \frac{d y}{d x} )\n(c) determine whether the lines tangent to the curve at the ( y ) intercept of\nthe curve are parallel. show the analysis that leads to your conclusion.

1. consider the curve defined by ( x^{2}+x y + 2 y^{2}=32 )\n(a) find the ( y ) intercepts of the curve defined above.\n(b) write an expression for the slope of the curve at the point ( (x, y) ), that is\nfind ( \frac{d y}{d x} )\n(c) determine whether the lines tangent to the curve at the ( y ) intercept of\nthe curve are parallel. show the analysis that leads to your conclusion.

Answer

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.