The combined equation of the tangent and norm | vedclass

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(D) Given curve is $xy = 100$. Differentiating with respect to $x$,we get $y + x \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = -\frac{y}{x}$.At point $(5, 20

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None of these

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(D) Given curve is $xy = 100$. Differentiating with respect to $x$,we get $y + x \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = -\frac{y}{x}$.At point $(5, 20)$,the slope of the tangent $m_1 = -\frac{20}{5} = -4$.The equation of the tangent is $y - 20 = -4(x - 5) \implies y - 20 = -4x + 20 \implies 4x + y - 40 = 0$.The slope of the normal $m_2 = -\frac{1}{m_1} = \frac{1}{4}$.The equation of the normal is $y - 20 = \frac{1}{4}(x - 5) \implies 4y - 80 = x - 5 \implies x - 4y + 75 = 0$.The combined equation is $(4x + y - 40)(x - 4y + 75) = 0$.Expanding this: $4x^2 - 16xy + 300x + xy - 4y^2 + 75y - 40x + 160y - 3000 = 0$.Simplifying: $4x^2 - 15xy - 4y^2 + 260x + 235y - 3000 = 0$.Comparing this with the given options,none of them match the result.

The combined equation of the tangent and normal to the curve $xy = 100$ at the point $(5, 20)$ is . . . . . . .

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