Question

The family of curves for which the area of the triangle formed by the x-axis, the tangent drawn at any point on the curve and radius vector of the point of tangency is constant equal to $2a^2$, is given by

• A. $x=cy\pm \frac{a^2}{y}$
• B. $y=cx\pm \frac{a^2}{y}$
• C. $x^2\pm a{y}^2=cy$
• D. $a^2{x}^2\pm y^2=cy$

Answer 1

Answer: (A)

Tangent drawn at any point P (x, y) is
$Y-y=\frac{dy}{dx}\left(X-x\right)...\left(i\right)$
The triangle , whose area is given is ${}_{\Delta }OPT$ (see the shaded region in the adjacent figure) If coordinates of T are $\left(X,0\right)$ then

$X=x-y\frac{dx}{dy}$ (Put $Y=0$ in $\left(i\right)$) $\therefore$ area of ${}_{\Delta }OPT$
$=\frac{1}{2}\left|X\cdot y\right|=2a^2$ (given)
$\therefore \frac{1}{2}\left|xy-y^2\frac{dx}{dy}\right|=2a^2\Rightarrow xy-y^2\frac{dx}{dy}=\pm 2a^2$
$\Rightarrow \frac{dx}{dy}-\frac{x}{y}=\pm \frac{2a^2}{y^2}$, which is linear in x
The integrating factor is $I.F.=e^{\int -\frac{1}{y}dy}{e}^{-logy}=\frac{1}{y}$
The solution is $x\left(\frac{1}{y}\right)=\int \pm \frac{2a^2}{y^2}\left(\frac{1}{y}\right)dy+c$
$\Rightarrow \frac{x}{y}=\pm \frac{2a^2{y}^{-2}}{-2}+\Rightarrow x=cy\pm \frac{a^2}{y}$