Question

. What is the orthogonal trajectories of the family of circles x2+y2=a2 ehere a is the parameter

(A) x=cy

(B) x2=cy2

(C) y=cx+b

(D) all of these​

Answer 1

Answer:

all of these --------------

Answer 2

Answer:

The orthogonal trajectories of the family of circles x2 + y2 = a2 is (A) x=cy

Explanation:

Given:

Equation of circle : $x^2+y^2=a^2$

To find:

Orthogonal trajectories

Step 1:

The given equation is $x^2+y^2=a^2$

Differentiating w.r.t. x, we get

$2x+2y\frac{dy}{dx} = 0$

$2y\frac{dy}{dx} = -2x$

$\frac{dy}{dx} = \frac{ -2x}{2y}$

$\frac{dy}{dx} = -\frac{x}{y}$

Step 2:

For orthogonal trajectories,

$\frac{dy}{dx} = \frac{-1}{\frac{dy}{dx} }$

Substituting the value of dy/dx

$\frac{dy}{dx} = \frac{-1}{\frac{-x}{y} }$

$\frac{dy}{dx} =\frac{y}{x}$

Therefore,

$\frac{dx}{x} =\frac{dy}{y} ...(i)$

Step 3:

Integrating equation (i), we get

$\int{\frac{dx}{x} } = \int{\frac{dy}{y}$

$log(x)=log(y)+log(c)$, where c is the integration constant.

$log(x) = log (cy)$

$x=cy$

Therefore, the orthogonal trajectory is x = cy.

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