The equation of the curve passing through the point $(1, 1)$ and having slope $\frac{2 a y}{x (y - a)}$ is

y^{a} . x^{2 a} = e^{y}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

y^{a} . x^{2 a} = e^{y - 1}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

x^{a} . y^{2 a} = e^{y}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $y^{a} . x^{2 a} = e^{y - 1}$
We have, $\frac{d y}{d x} = \frac{2 a y}{x (y - a)}$
$\Rightarrow \frac{y - a}{y} d y = \frac{2 a}{x} d x$
On integrating both sides, we get
$log | y | - y = - 2 a log | x | + log c \Rightarrow y^{a} . x^{2 a} = {c e}^{y}$
Since, the curve passes through $(1, 1)$ therefore
$1 = c e \Rightarrow c = \frac{1}{e}$
So, the equation of the curve is
$y^{a} . x^{2 a} = e^{y - 1}$

Suggest Corrections

Similar questions

\frac{2 y}{x}

x = e^{y + e^{y + e^{y + e^{y + \dots \infty}}}}

\frac{d x}{d y} - \frac{x log x}{1 + log x} = \frac{e^{y}}{1 + log x}

y (1) = 0

m

e^{y} = 1 + x^{2}

x = e^{y + e^{y + e^{y + e^{y + \dots \infty}}}}

Join BYJU'S Learning Program