Search any question & find its solution
Question:
Answered & Verified by Expert
The general solution of $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{ax}+\mathrm{h}}{\mathrm{by}+\mathrm{k}}$ represents a circle
only when $\quad$
Options:
only when $\quad$
Solution:
2070 Upvotes
Verified Answer
The correct answer is:
$a=-b \neq 0$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{ax}+\mathrm{h}}{\mathrm{by}+\mathrm{k}}$
$\Rightarrow(\mathrm{by}+\mathrm{k}) \mathrm{dy}=(\mathrm{ax}+\mathrm{h}) \mathrm{d} \mathrm{x}$
Intergrating, we get $\int(\mathrm{by}+\mathrm{k}) \mathrm{dy}=\int(\mathrm{ax}+\mathrm{h}) \mathrm{dx}$
$\Rightarrow \frac{\mathrm{by}^{2}}{2}+\mathrm{ky}=\frac{\mathrm{ax}^{2}}{2}+\mathrm{hx}$
$\Rightarrow b y^{2}+2 k y=a x^{2}+2 h x$
$\Rightarrow a x^{2}-b y^{2}+2 h x-2 k y=0$
This represents circle only when $\mathrm{a}=-\mathrm{b} \neq 0$.
$\Rightarrow(\mathrm{by}+\mathrm{k}) \mathrm{dy}=(\mathrm{ax}+\mathrm{h}) \mathrm{d} \mathrm{x}$
Intergrating, we get $\int(\mathrm{by}+\mathrm{k}) \mathrm{dy}=\int(\mathrm{ax}+\mathrm{h}) \mathrm{dx}$
$\Rightarrow \frac{\mathrm{by}^{2}}{2}+\mathrm{ky}=\frac{\mathrm{ax}^{2}}{2}+\mathrm{hx}$
$\Rightarrow b y^{2}+2 k y=a x^{2}+2 h x$
$\Rightarrow a x^{2}-b y^{2}+2 h x-2 k y=0$
This represents circle only when $\mathrm{a}=-\mathrm{b} \neq 0$.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.