The family of curves whose x and y intercepts | vedclass

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Question

(A) Let the point on the curve be $(x, y)$.The equation of the tangent at $(x, y)$ is given by $Y - y = \frac{dy}{dx}(X - x)$,where $(X, Y)$ are the c

Vedclass · Accepted Answer

$xy = C$

Vedclass · Answer

(A) Let the point on the curve be $(x, y)$.The equation of the tangent at $(x, y)$ is given by $Y - y = \frac{dy}{dx}(X - x)$,where $(X, Y)$ are the coordinates of any point on the tangent.The $x$-intercept is found by setting $Y = 0$: $-y = \frac{dy}{dx}(X - x) \Rightarrow X = x - y \frac{dx}{dy}$.The $y$-intercept is found by setting $X = 0$: $Y - y = \frac{dy}{dx}(-x) \Rightarrow Y = y - x \frac{dy}{dx}$.According to the problem,the $x$-intercept is $2x$ and the $y$-intercept is $2y$.So,$x - y \frac{dx}{dy} = 2x \Rightarrow -y \frac{dx}{dy} = x \Rightarrow -\frac{dx}{x} = \frac{dy}{y}$.Integrating both sides: $-\int \frac{dx}{x} = \int \frac{dy}{y} \Rightarrow -\ln|x| = \ln|y| + \ln|C|$.This simplifies to $\ln|y| + \ln|x| = \ln|C|$,which gives $xy = C$.

The family of curves whose $x$ and $y$ intercepts of a tangent at any point are respectively double the $x$ and $y$ coordinates of that point is

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