The normal to the curve y(x-2)(x-3)=x+6 at th | vedclass

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(B) The given equation of the curve is $y(x-2)(x-3)=x+6$. At the $Y$-axis,$x=0$. Substituting $x=0$ into the equation: $y(0-2)(0-3)=0+6 \Rightarrow y(

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$\left(\frac{1}{2}, \frac{1}{2}\right)$

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(B) The given equation of the curve is $y(x-2)(x-3)=x+6$. At the $Y$-axis,$x=0$. Substituting $x=0$ into the equation: $y(0-2)(0-3)=0+6 \Rightarrow y(-2)(-3)=6 \Rightarrow 6y=6 \Rightarrow y=1$. So,the point of intersection is $(0, 1)$. Now,rewrite the equation as $y = \frac{x+6}{x^2-5x+6}$. Differentiating with respect to $x$ using the quotient rule: $\frac{dy}{dx} = \frac{(x^2-5x+6)(1) - (x+6)(2x-5)}{(x^2-5x+6)^2}$. At $x=0$,the slope of the tangent $m_t = \frac{(0-0+6)(1) - (0+6)(0-5)}{(0-0+6)^2} = \frac{6 - (-30)}{36} = \frac{36}{36} = 1$. The slope of the normal $m_n = -\frac{1}{m_t} = -\frac{1}{1} = -1$. The equation of the normal at $(0, 1)$ is $y - 1 = -1(x - 0)$,which simplifies to $y = -x + 1$ or $x + y = 1$. Checking the options,for $(\frac{1}{2}, \frac{1}{2})$,we have $\frac{1}{2} + \frac{1}{2} = 1$,which satisfies the equation. Thus,the normal passes through $(\frac{1}{2}, \frac{1}{2})$.

The normal to the curve $y(x-2)(x-3)=x+6$ at the point,where the curve intersects the $Y$-axis,passes through the point

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