Question

Find the orthogonal trajectory of the family of the curve $x^2 + y^2 = ax$.

Answer 1

x^2+y^2 = k,y

Answer 2

We are given the family

$$x^2+y^2 = ax,$$

where $a$ is a parameter.

Step 1. Find the differential equation of the family

Differentiate both sides with respect to $x$:

$$\frac{d}{dx}(x^2+y^2)=\frac{d}{dx}(ax).$$

This gives:

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

Since from the original family $a = \frac{x^2+y^2}{x}$ (assuming $x\ne0$), substitute into the differentiated equation:

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

Multiply both sides by $x$:

$$2x^2 + 2xy\frac{dy}{dx} = x^2+y^2.$$

Solve for $\frac{dy}{dx}$:

$$2xy\frac{dy}{dx} = y^2 - x^2 \quad\Longrightarrow\quad \frac{dy}{dx} = \frac{y^2-x^2}{2xy}.$$

Step 2. Write the differential equation for the orthogonal trajectories

For orthogonal trajectories, the slope at any point must be the negative reciprocal. Thus, if

$$m = \frac{y^2-x^2}{2xy},$$

then the slope of the orthogonal trajectories is

$$\frac{dy}{dx} = -\frac{1}{m} = -\frac{2xy}{y^2-x^2}.$$

Step 3. Solve the differential equation using substitution

Let

$$v=\frac{y}{x} \quad\Longrightarrow\quad y = vx \quad\text{and}\quad \frac{dy}{dx} = v + x\frac{dv}{dx}.$$

Substitute into the orthogonal differential equation:

$$v + x\frac{dv}{dx} = -\frac{2x(vx)}{(v^2x^2-x^2)} = -\frac{2v x^2}{x^2(v^2-1)} = -\frac{2v}{v^2-1}.$$

This simplifies to:

$$x\frac{dv}{dx} = -\frac{2v}{v^2-1} - v = -v\left[\frac{2}{v^2-1}+1\right].$$

Combine the terms in the bracket:

$$\frac{2}{v^2-1}+1 = \frac{2+ (v^2-1)}{v^2-1} = \frac{v^2+1}{v^2-1}.$$

Thus,

$$x\frac{dv}{dx} = -v\frac{v^2+1}{v^2-1}.$$

Separate the variables:

$$\frac{v^2-1}{v(v^2+1)}\,dv = -\frac{dx}{x}.$$

Step 4. Integrate both sides

We now integrate:

$$\int \frac{v^2-1}{v(v^2+1)}\,dv = -\int \frac{dx}{x}.$$

Perform partial fractions on the integrand. Write:

$$\frac{v^2-1}{v(v^2+1)} = \frac{A}{v}+\frac{Bv+C}{v^2+1}.$$

Clearing the denominators:

$$v^2-1 = A(v^2+1) + (Bv+C)v = (A+B)v^2+ Cv+A.$$

Matching coefficients:

• Coefficient of $v^2$: $A+B=1$.
• Coefficient of $v$: $C=0$.
• Constant term: $A=-1$.

Thus,

$$A=-1,\quad B=2,\quad C=0.$$

So the integrand becomes:

$$\frac{-1}{v} + \frac{2v}{v^2+1}.$$

Integrate term by term:

$$\int \left(-\frac{1}{v} + \frac{2v}{v^2+1}\right)dv = -\ln|v| + \ln|v^2+1| + C_1 = \ln\left|\frac{v^2+1}{v}\right|+C_1.$$

The right side integrates to:

$$-\int \frac{dx}{x} = -\ln|x|+C_2.$$

Combining the constants, we have:

$$\ln\left|\frac{v^2+1}{v}\right| = -\ln|x| + C.$$

Step 5. Express the answer in terms of $x$ and $y$

Recall that $v=\frac{y}{x}$. Notice that:

$$\frac{v^2+1}{v} = \frac{\left(\frac{y}{x}\right)^2+1}{\frac{y}{x}} = \frac{\frac{y^2}{x^2}+1}{\frac{y}{x}} = \frac{\frac{x^2+y^2}{x^2}}{\frac{y}{x}} = \frac{x^2+y^2}{xy}.$$

Thus, the integrated result becomes:

$$\ln\left|\frac{x^2+y^2}{xy}\right| = -\ln|x| + C.$$

Exponentiating both sides gives:

$$\frac{x^2+y^2}{xy} = \frac{K}{x} \quad\text{or}\quad x^2+y^2 = K\; y,$$

where $K$ is an arbitrary constant.

Final Answer:

The orthogonal trajectories of the family

$$x^2+y^2=ax$$

are given by

$$x^2+y^2 = k\,y,$$

where $k$ is an arbitrary constant.