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.