Question

the graph of the equation ( x^{2}+xy + y^{2}=5 ) is an ellipse lying obliquely in the plane, as illustrated in the figure below.
a. compute ( \frac{dy}{dx} ).
b. the ellipse has two horizontal tangents. find an equation of the upper one.
the upper horizontal tangent line is defined by the equation ( y=square )
c. the ellipse has two vertical tangents. find an equation of the rightmost one.
the rightmost vertical tangent line is defined by the equation ( x=square )
d. find the point at which the rightmost vertical tangent line touches the ellipse.
the rightmost vertical tangent line touches the ellipse at the
hint: the horizontal tangent is of course characterized by ( \frac{dy}{dx}=0 ). to find the vertical tangent use symmetry, or solve ( \frac{dx}{dy}=0 ).

Explanation:

Step1: Differentiate implicitly

Given \(x^{2}+xy + y^{2}=5\). Differentiate both sides with respect to \(x\).
Using the sum rule \((u + v+w)^\prime=u^\prime + v^\prime+w^\prime\), where \(u = x^{2}\), \(v=xy\), \(w = y^{2}\).
For \(u=x^{2}\), \(u^\prime = 2x\).
For \(v = xy\), use the product rule \((uv)^\prime=u^\prime v+uv^\prime\), here \(u = x\), \(v = y\), so \(v^\prime=y + x\frac{dy}{dx}\).
For \(w=y^{2}\), use the chain - rule \(\frac{d}{dx}(y^{2})=2y\frac{dy}{dx}\).
So, \(2x+(y + x\frac{dy}{dx})+2y\frac{dy}{dx}=0\).

Step2: Solve for \(\frac{dy}{dx}\)

Group the terms with \(\frac{dy}{dx}\):
\(2x + y+(x + 2y)\frac{dy}{dx}=0\).
\((x + 2y)\frac{dy}{dx}=-2x - y\).
\(\frac{dy}{dx}=\frac{-2x - y}{x + 2y}\).

Answer:

\(\frac{dy}{dx}=\frac{-2x - y}{x + 2y}\)