Question

Tangent to a curve intersects the y-axis at a point $\mathrm{P}$. A line perpendicular to this tangent through $\mathrm{P}$ passes through the point $\left(1,0\right)$. The differential equation of the curve is

• A. $y \frac{dy}{dx} - x ( \frac {dy}{dx} ) ^2$
• B. $y \frac{dy}{dx} + x ( \frac {dy}{dx} ) ^2$
• C. $\mathrm{y}\frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\mathrm{y}}+\mathrm{x}=1$
• D. None of these

Answer 1

Answer: (A)

Equation of tangent at the point
$\mathrm{R}(\mathrm{x}, \mathrm{f}(\mathrm{x}))$ is,
$\mathrm{Y}-\mathrm{f}(\mathrm{x})=\mathrm{f}^{\prime}(\mathrm{x})(\mathrm{X}-\mathrm{x})$
Coordinates of the point are $\mathrm{P}\left(0, \mathrm{f}(\mathrm{x})-\mathrm{xf}^{\prime}(\mathrm{x})\right)$
The slope of the perpendicular line through $\mathrm{P}$
is $\frac{\mathrm{f}(\mathrm{x})-\mathrm{xf}^{\prime}(\mathrm{x})}{-1}=-\frac{1}{\mathrm{f}^{\prime}(\mathrm{x})}$
$\mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{x}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}=1$
is the differential equation.