Find the equation of the normal to the curve | vedclass

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(D) The given curve is $y = \frac{x-7}{(x-2)(x-3)}$. The curve cuts the $X$-axis where $y = 0$. Setting $y = 0$,we get $x - 7 = 0$,so $x = 7$. Thus,th

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$20x + y - 140 = 0$

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(D) The given curve is $y = \frac{x-7}{(x-2)(x-3)}$. The curve cuts the $X$-axis where $y = 0$. Setting $y = 0$,we get $x - 7 = 0$,so $x = 7$. Thus,the point of intersection is $P(7, 0)$.To find the slope of the tangent,we differentiate $y$ with respect to $x$ using the quotient rule:$\frac{dy}{dx} = \frac{[(x-2)(x-3)] \cdot 1 - (x-7) \cdot \frac{d}{dx}[(x-2)(x-3)]}{[(x-2)(x-3)]^2}$$\frac{dy}{dx} = \frac{(x^2 - 5x + 6) - (x-7)(2x - 5)}{(x-2)^2(x-3)^2}$At $x = 7$,the denominator is $(7-2)^2(7-3)^2 = 5^2 \cdot 4^2 = 25 \cdot 16 = 400$. The numerator is $(49 - 35 + 6) - (0) = 20$.So,$\left. \frac{dy}{dx} \right|_{x=7} = \frac{20}{400} = \frac{1}{20}$.The slope of the normal is $m_n = -\frac{1}{dy/dx} = -20$.The equation of the normal at $(7, 0)$ is $y - 0 = -20(x - 7)$,which simplifies to $y = -20x + 140$,or $20x + y - 140 = 0$.

Find the equation of the normal to the curve $y = \frac{x-7}{(x-2)(x-3)}$ at the point where it cuts the $X$-axis.

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