Question

From the differential equation of the family of hyperbolas having foci on \(x\)-axis and centre at origin.

Answer 1

The correct answer is:\(xyy''+x(y')^2-yy'=0\)
The equation of the family of hyperbolas with the centre at origin and foci along the xaxis is:
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1...(1)\)

Differentiating equation(1)with respect to \(x\),we get:
\(\frac{2x}{a^2}+\frac{2yy'}{b^2}=0\)
\(⇒\frac{x}{a^2}+\frac{yy'}{b^2}=0...(2)\)
Again,differentiating with respect to \(x\),we get:
\(\frac{1}{a^2}-\frac{y'.y'+y.y''}{b^2}=0\)
Substituting the value of \(\frac{1}{a^2}\) in equation (2), we get:
\(\frac{x}{b^2}[((y')2+yy'')]+\frac{yy'}{b^2}=0\)
\(⇒-x(y')^2-xyy''+yy'=0\)
\(⇒xyy''+x(y')^2-yy'=0\)
This is the required differential equation.