A normal is drawn at a point P(x, y) of a cur | vedclass

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(A) Let the point $P$ be $(x, y)$. The slope of the tangent at $P$ is $\frac{dy}{dx}$. The slope of the normal at $P$ is $-\frac{dx}{dy}$.The equation

Vedclass · Accepted Answer

$x^2 + y^2 = k^2$

Vedclass · Answer

(A) Let the point $P$ be $(x, y)$. The slope of the tangent at $P$ is $\frac{dy}{dx}$. The slope of the normal at $P$ is $-\frac{dx}{dy}$.The equation of the normal at $P(x, y)$ is $Y - y = -\frac{dx}{dy}(X - x)$.To find the point $Q$ where the normal meets the $X$-axis,set $Y = 0$: $-y = -\frac{dx}{dy}(X - x) \implies X - x = y \frac{dy}{dx} \implies X = x + y \frac{dy}{dx}$.Thus,$Q = (x + y \frac{dy}{dx}, 0)$.The length $PQ$ is given by $l(PQ)^2 = (X - x)^2 + (0 - y)^2 = k^2$.Substituting $X - x = y \frac{dy}{dx}$,we get $(y \frac{dy}{dx})^2 + y^2 = k^2$.$y^2 ((\frac{dy}{dx})^2 + 1) = k^2 \implies (\frac{dy}{dx})^2 = \frac{k^2 - y^2}{y^2} = \frac{k^2 - y^2}{y^2}$.$\frac{dy}{dx} = \pm \frac{\sqrt{k^2 - y^2}}{y}$.Separating variables: $\int \frac{y}{\sqrt{k^2 - y^2}} dy = \pm \int dx$.Let $u = k^2 - y^2$,then $du = -2y dy$. The integral becomes $-\frac{1}{2} \int u^{-1/2} du = -\sqrt{u} = -\sqrt{k^2 - y^2}$.So,$-\sqrt{k^2 - y^2} = \pm x + C$.Since the curve passes through $(0, k)$,we have $-\sqrt{k^2 - k^2} = 0 + C \implies C = 0$.Thus,$\sqrt{k^2 - y^2} = |x| \implies k^2 - y^2 = x^2 \implies x^2 + y^2 = k^2$.

$A$ normal is drawn at a point $P(x, y)$ of a curve $y=f(x)$. The normal meets the $X$-axis at $Q$. If the length $l(PQ) = k$,where $k$ is a constant,find the equation of the curve passing through $(0, k)$.

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