The length of the perpendicular from the orig | vedclass

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(B) Given the curve $x^{2}+2xy-3y^{2}=0$. Differentiating with respect to $x$: $2x + 2y + 2x\frac{dy}{dx} - 6y\frac{dy}{dx} = 0$. At point $(2,2)$: $2

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$2\sqrt{2}$

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(B) Given the curve $x^{2}+2xy-3y^{2}=0$. Differentiating with respect to $x$: $2x + 2y + 2x\frac{dy}{dx} - 6y\frac{dy}{dx} = 0$. At point $(2,2)$: $2(2) + 2(2) + 2(2)\frac{dy}{dx} - 6(2)\frac{dy}{dx} = 0$. $4 + 4 + 4\frac{dy}{dx} - 12\frac{dy}{dx} = 0 \implies 8 - 8\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = 1$. The slope of the tangent at $(2,2)$ is $m_{T} = 1$. The slope of the normal at $(2,2)$ is $m_{N} = -\frac{1}{m_{T}} = -1$. The equation of the normal at $(2,2)$ is $(y - 2) = -1(x - 2) \implies y - 2 = -x + 2 \implies x + y - 4 = 0$. The perpendicular distance from the origin $(0,0)$ to the line $x + y - 4 = 0$ is given by $d = \frac{|ax_{0} + by_{0} + c|}{\sqrt{a^{2} + b^{2}}}$. $d = \frac{|0 + 0 - 4|}{\sqrt{1^{2} + 1^{2}}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.

The length of the perpendicular from the origin,on the normal to the curve $x^{2}+2xy-3y^{2}=0$ at the point $(2,2)$ is

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