The curve,with the property that the projecti | vedclass

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(A) Let $P(x, y)$ be a point on the curve. Let $PM$ be the ordinate,where $M$ is on the $x$-axis. Let the normal at $P$ meet the $x$-axis at $T$. Let

Vedclass · Accepted Answer

$x + a \ln \left( \sqrt{y^2 - a^2} + y \right) = c$

Vedclass · Answer

(A) Let $P(x, y)$ be a point on the curve. Let $PM$ be the ordinate,where $M$ is on the $x$-axis. Let the normal at $P$ meet the $x$-axis at $T$. Let $	heta$ be the angle the tangent makes with the $x$-axis. Then the normal makes an angle $\pi - 	heta$ with the $x$-axis. The projection of the ordinate $PM$ on the normal $PT$ is $PN$. In the right-angled triangle $	riangle PMN$,the angle $\angle PMN = 	heta$. Thus,$PN = PM \cos 	heta = y \cos 	heta$. Given $PN = a$,we have $y \cos 	heta = a$. Since $\cos 	heta = \frac{1}{\sqrt{1 + 	an^2 	heta}}$ and $	an 	heta = \frac{dy}{dx} = y_1$,we get $y \frac{1}{\sqrt{1 + y_1^2}} = a$. Squaring both sides,$y^2 = a^2(1 + y_1^2)$,which gives $y_1^2 = \frac{y^2 - a^2}{a^2}$. Thus,$\frac{dy}{dx} = \frac{\sqrt{y^2 - a^2}}{a}$. Separating variables,$\int \frac{a \, dy}{\sqrt{y^2 - a^2}} = \int dx$. Integrating,we get $a \ln |y + \sqrt{y^2 - a^2}| = x + c$.

The curve,with the property that the projection of the ordinate on the normal is constant and has a length equal to $a$,is

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