The equation of the family of curves for whic | vedclass

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(A) The length of the normal to a curve is given by $L = |y| \sqrt{1 + (\frac{dy}{dx})^2}$.Given that the length of the normal is equal to the radius

Vedclass · Accepted Answer

${y^2} \pm {x^2} = k$

Vedclass · Answer

(A) The length of the normal to a curve is given by $L = |y| \sqrt{1 + (\frac{dy}{dx})^2}$.Given that the length of the normal is equal to the radius vector $r = \sqrt{x^2 + y^2}$,we have:$|y| \sqrt{1 + (\frac{dy}{dx})^2} = \sqrt{x^2 + y^2}$Squaring both sides:$y^2 (1 + (\frac{dy}{dx})^2) = x^2 + y^2$$y^2 + y^2 (\frac{dy}{dx})^2 = x^2 + y^2$$y^2 (\frac{dy}{dx})^2 = x^2$Taking the square root on both sides:$y \frac{dy}{dx} = \pm x$Separating the variables:$y \, dy = \pm x \, dx$Integrating both sides:$\int y \, dy = \pm \int x \, dx$$\frac{y^2}{2} = \pm \frac{x^2}{2} + C$$y^2 = \pm x^2 + 2C$Let $2C = k$,then the equation is ${y^2} \mp {x^2} = k$,which is equivalent to ${y^2} \pm {x^2} = k$.

The equation of the family of curves for which the length of the normal is equal to the radius vector is

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