The equation of a curve whose normal at any p | vedclass

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(A) Let the point on the curve be $(x, y)$. The slope of the tangent is $\frac{dy}{dx}$.The slope of the normal is $-\frac{dx}{dy}$.According to the p

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$1$

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(A) Let the point on the curve be $(x, y)$. The slope of the tangent is $\frac{dy}{dx}$.The slope of the normal is $-\frac{dx}{dy}$.According to the problem,the slope of the normal is equal to the ordinate $y$,so $-\frac{dx}{dy} = y$.This implies $dx = -y \, dy$.Integrating both sides,we get $x = -\frac{y^2}{2} + C$.Since the curve passes through $(1, -1)$,we substitute these values: $1 = -\frac{(-1)^2}{2} + C$,which gives $1 = -\frac{1}{2} + C$,so $C = \frac{3}{2}$.Thus,the equation is $x = -\frac{y^2}{2} + \frac{3}{2}$,which simplifies to $2x = -y^2 + 3$,or $2x = 1(3 - y^2)$.Comparing this with $2x = k(3 - y^2)$,we find $k = 1$.

The equation of a curve whose normal at any point has a slope which is the same as the ordinate and which passes through $(1, -1)$ is $2x = k(3 - y^2)$. Then $k$ is:

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