The length of the perpendicular drawn from th | vedclass

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(C) 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)

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$2\sqrt{2}$ units

Vedclass · Answer

(C) 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 $m = 1$. The slope of the normal is $m' = -\frac{1}{m} = -1$. 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 length of the perpendicular from the origin $(0,0)$ to the line $x + y - 4 = 0$ is given by $d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$. $d = \frac{|1(0) + 1(0) - 4|}{\sqrt{1^2 + 1^2}} = \frac{|-4|}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$ units.

The length of the perpendicular drawn 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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