$\text { Find the orthogonal trajectories of the family } 3 x ^ { 2 } + 6 y ^ { 2 } = C \text {. }$
A) $6 y ^ { 2 } - 3 x ^ { 2 } = k$
B) $| y | ^ { 3 } = k | x | ^ { 6 }$
C) $| y | ^ { 6 } = k | x | ^ { - 6 }$
D) $6 y ^ { 2 } + 3 x ^ { 2 } = k$
E) $| y | ^ { 3 } = k | x | ^ { - 6 }$

Question

$\text { Find the orthogonal trajectories of the family } 3 x ^ { 2 } + 6 y ^ { 2 } = C \text {. }$ 
A) $6 y ^ { 2 } - 3 x ^ { 2 } = k$
B) $| y | ^ { 3 } = k | x | ^ { 6 }$
C) $| y | ^ { 6 } = k | x | ^ { - 6 }$
D) $6 y ^ { 2 } + 3 x ^ { 2 } = k$
E) $| y | ^ { 3 } = k | x | ^ { - 6 }$

Quizplus · Accepted Answer

To find the orthogonal trajectories, we need to differentiate the given equation with respect to y to get

$$6y\frac{dy}{dx}+6x=-\frac{d}{dx} \left(\frac{C}{3x^2+6y^2}\right)\frac{dx}{dy}$$

Then we simplify to

$$\frac{dy}{dx}=-\frac{3x^2+6y^2}{6xy}=\frac{1}{2}\left(-\frac{x}{y}-\frac{y}{x}\right) $$

The condition of perpendicularity between the family of curves and its orthogonal trajectories gives us

$$\frac{dy}{dx}=2\left(\frac{x}{y}+\frac{y}{x}\right) $$

Simplifying this by clearing denominators gives us

$$2x^2y+2y^3=K$$

This is the equation of the family of orthogonal trajectories, which is equivalent to

$$|y|^3=k|x|^6$$

Therefore, the answer is B.

Find the orthogonal trajectories of the family 3x2+6y2=C. \text { Find the orthogonal trajectories of the family } 3 x ^ { 2 } + 6 y ^ { 2 } = C \text {. } Find the orthogonal trajectories of the family 3x2+6y2=C.

$\text { Find the orthogonal trajectories of the family } 3 x ^ { 2 } + 6 y ^ { 2 } = C \text {. }$