The equation of a curve passing through (1, 0 | vedclass

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Question

(A) Let the point $P$ be $(x, y)$. The slope of the tangent at $P$ is $m = \frac{dy}{dx}$. The slope of the normal at $P$ is $-\frac{1}{m} = -\frac{dx

Vedclass · Accepted Answer

$x^2 + y^2 = x^4$

Vedclass · Answer

(A) Let the point $P$ be $(x, y)$. The slope of the tangent at $P$ is $m = \frac{dy}{dx}$. The slope of the normal at $P$ is $-\frac{1}{m} = -\frac{dx}{dy}$.The equation of the normal at $P(x, y)$ is $Y - y = -\frac{dx}{dy} (X - x)$.To find the $x$-intercept,set $Y = 0$:$-y = -\frac{dx}{dy} (X - x) \implies X - x = y \frac{dy}{dx} \implies X = x + y \frac{dy}{dx}$.The abscissa of $P$ is $x$. The $x$-intercept of the normal is $X = x + y \frac{dy}{dx}$.The radius vector of $P$ is $\sqrt{x^2 + y^2}$,so its square is $x^2 + y^2$.According to the problem,the product of the abscissa and the $x$-intercept equals twice the square of the radius vector:$x \left( x + y \frac{dy}{dx} \right) = 2(x^2 + y^2)$.$x^2 + xy \frac{dy}{dx} = 2x^2 + 2y^2 \implies xy \frac{dy}{dx} = x^2 + 2y^2$.This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.$x(vx) (v + x \frac{dv}{dx}) = x^2 + 2(vx)^2 \implies x^2 v (v + x \frac{dv}{dx}) = x^2(1 + 2v^2)$.$v^2 + vx \frac{dv}{dx} = 1 + 2v^2 \implies vx \frac{dv}{dx} = 1 + v^2$.$\frac{v}{1 + v^2} dv = \frac{1}{x} dx$.Integrating both sides: $\frac{1}{2} \ln(1 + v^2) = \ln x + C \implies \ln(1 + v^2) = \ln x^2 + 2C \implies 1 + v^2 = k x^2$,where $k = e^{2C}$.Substituting $v = \frac{y}{x}$: $1 + \frac{y^2}{x^2} = k x^2 \implies x^2 + y^2 = k x^4$.The curve passes through $(1, 0)$,so $1^2 + 0^2 = k(1)^4 \implies k = 1$.Thus,the equation is $x^2 + y^2 = x^4$.

The equation of a curve passing through $(1, 0)$ for which the product of the abscissa of a point $P$ and the intercept made by a normal at $P$ on the $x$-axis equals twice the square of the radius vector of the point $P$ is

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