The normal to a curve at P(x, y) meets the x- | vedclass

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(C) 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

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a hyperbola

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(C) 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 $G$ where the normal meets the $x$-axis, set $Y = 0$:$-y = -\frac{dx}{dy}(X - x) \implies y \frac{dy}{dx} = X - x \implies X = x + y \frac{dy}{dx}$.The point $G$ is $(x + y \frac{dy}{dx}, 0)$.The distance of $G$ from the origin is $|x + y \frac{dy}{dx}|$. Given that this distance is twice the abscissa of $P$ (which is $x$), we have:$x + y \frac{dy}{dx} = 2x \implies y \frac{dy}{dx} = x$.Integrating both sides with respect to $x$:$\int y \, dy = \int x \, dx \implies \frac{y^2}{2} = \frac{x^2}{2} + C \implies y^2 - x^2 = 2C$.This represents a rectangular hyperbola.

The normal to a curve at $P(x, y)$ meets the $x$-axis at $G$. If the distance of $G$ from the origin is twice the abscissa of $P$, then the curve is

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