Question

Form a differential equation of family of parabolas with focus origin and axis of symmetry along the \(\mathrm{x}\)-axis.

Short answer

Question: Write the differential equation representing the family of parabolas with their focus at origin and axis of symmetry along the x-axis. Answer: The differential equation for the given family of parabolas is \(2yy' = \frac{y^2}{x}\).

Step-by-step solution

Find the general equation of a parabola with given conditions

Any parabola with focus at (0,0) and axis of symmetry along the x-axis can be represented by the equation \((y-k)^2=4a(x-h)\), where \((h,k)\) is the vertex of the parabola. In this case, we want all parabolas to have their vertex on the x-axis which means the y-coordinate of the vertex will be 0. Therefore, the equation becomes \((y-0)^2=4a(x-h)\) or \(y^2=4a(x-h)\).

Write the equation in terms of the arbitrary constant a

We know that the vertex is on the x-axis, so we can write \(y^2=4a(x-h)\) as \(y^2=4a(x-0)\). This allows us to rewrite the equation as \(y^2=4ax\).

Differentiate the equation with respect to x

Now, we differentiate the equation \(y^2=4ax\) with respect to x. Using the chain rule, we have \(2yy'=4a\).

Eliminate the arbitrary constant a

The arbitrary constant a is present in the equation after differentiation. To eliminate it, we will use the original equation \(y^2=4ax\). We can write \(a=\frac{y^2}{4x}\). Substitute this expression for a in the equation from step 3: \(2yy' = 4\left(\frac{y^2}{4x}\right)\).

Simplify and obtain the final differential equation

By simplifying the expression from step 4, we get: \(2yy' = y^2\frac{4}{4x}\) or \(2yy' = y^2\frac{1}{x}\). So, the differential equation representing the family of parabolas with focus at origin and axis of symmetry along the x-axis is: \(2yy' = \frac{y^2}{x}\)