Part A. Find the orthogonal trajectories of the family of curves. (Use C for any needed constant.)x2 + 2y2 = 17k2Part B. Find the orthogonal trajectories of the family of curves. (Use C for any needed constant.)y2 = 7kx3

Answer:

Following are the solution to the question:

Step-by-step explanation:

The objective is to find the orthogonal trajectory of the family curves:

[tex]x^2+2y^2=17k^2[/tex]

The family curves are: [tex]x^2 + 2y^2=17k^2[/tex]  

Differentiate the equation with the respect of x.

[tex]\to x^2+2y^2=17k^2 \\\\\to 2x+4y \frac{dy}{dx} =0\\\\\to \frac{dy}{dx} =- \frac{2x}{4y}\\\\\to \frac{dy}{dx} =- \frac{x}{2y}[/tex]

The negative reciprocal is:

[tex]\to \frac{dy}{dx} =-\frac{1}{-\frac{x}{2y}}\\\\\to \frac{dy}{dx} = \frac{2y}{x}[/tex]

Integration of the above equation:

[tex]\to \frac{dy}{dx} = \frac{2y}{x} \\\\\to \int \frac{dy}{2y} = \int \frac{dx}{x}\\\\\to \frac{1}{2} \log y = \log x + \log C\\\\\to \log y = 2 \log x + 2\log C \\\\\to \log y = \log x^2 +\log C^2 \\\\\to y = Ax^2 \ \ \ \ \ \ \ \ \ (A=C^2)[/tex]

Hence, the orthogonal trajectory of the family of the curves [tex]x^2 + 2y^2 =17k^2 \ \ is \ \ y=Ax^2[/tex]

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