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.