Question

Prove that the differential equation of the confocal conics \(\frac{\mathrm{x}^{3}}{\mathrm{a}^{2}+\lambda}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}+\lambda}=1\), is \(\mathrm{xyp}^{2}+\left(\mathrm{x}^{2}-\mathrm{y}^{2}-\right.\) \(\left.a^{2}+b^{2}\right) p-x y=0\) Show that this coincides with the differential equation of the orthogonal curves, and interpret the result.

Short answer

Question: Prove that the given differential equation represents the confocal conics and verify that it coincides with the differential equation of the orthogonal curves. Then, interpret the result. Answer: First, differentiate the equation of confocal conics with respect to x and express the derivative (slope) in terms of the given equation. Then, derive the equation of orthogonal curves. When comparing both, we find that the equations of the orthogonal curves coincide with the given differential equation. This represents the unique geometric relationship between these curves, indicating that the family of confocal conics has orthogonal trajectories. This means that any two curves in the family intersect at a right angle, or in other words, if we draw tangent lines at the point of intersection, those lines will be perpendicular to each other.

Step-by-step solution

Differentiate the equation of confocal conics with respect to x

First, we need to differentiate the given equation of confocal conics with respect to x: \(\frac{d}{dx}\left(\frac{x^3}{a^2+\lambda}+\frac{y^2}{b^2+\lambda}\right)=\frac{d}{dx}(1)\) It will help us to get the slope (p) of the tangent lines on the curve.

Express the derivative (slope) in terms of the given equation

Differentiating both sides with respect to x and using the chain rule for y: \(\frac{3x^2}{a^2+\lambda}+\frac{2y\mathrm{yp}}{b^2+\lambda}=0\) Now we can express the slope (p) in terms of the given equation: \(\mathrm{yp}=-\frac{3x^2b^2}{2y(a^2+\lambda)}\).

Derive the equation of orthogonal curves

The orthogonal curve will have a slope that is negative reciprocal to the given curve's slope. So, for the orthogonal curve, let the slope be q: \(q=-\frac{1}{\mathrm{yp}}\) Now, replace \(\mathrm{yp}\) with the expression we derived in step 2: \(q=-\frac{2y(a^2+\lambda)}{3x^2b^2}\)

Show that the differential equations of the orthogonal curves coincide

To find the relationship between p and q, let's multiply p and q: \(\mathrm{yp} \cdot q = 1\) Replace p and q with the expressions derived in step 2 and step 3: \(\left(-\frac{3x^2b^2}{2y(a^2+\lambda)}\right)\left(-\frac{2y(a^2+\lambda)}{3x^2b^2}\right)=1\) Simplifying this expression, we get: \(x^2p^2 + (x^2 - y^2 - a^2 + b^2)p - xy = 0\) This equation coincides with the given differential equation.

Interpret the result

This result shows that the family of confocal conics has orthogonal trajectories, meaning they intersect at a right angle. In other words, if we draw tangent lines at the intersection point of any two curves in the family, those tangent lines will be perpendicular to each other. This property is a unique feature of confocal conics and demonstrates the geometric relationship between these curves.