Find the equation of the normal to the parabo | vedclass

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(A) Given the equation of the parabola is $y^2 = 8x$. Comparing this with $y^2 = 4ax$,we get $4a = 8$,so $a = 2$. The point given is $(x_1, y_1) = (2,

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$x + y = 6$

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(A) Given the equation of the parabola is $y^2 = 8x$. Comparing this with $y^2 = 4ax$,we get $4a = 8$,so $a = 2$. The point given is $(x_1, y_1) = (2, 4)$. Differentiating $y^2 = 8x$ with respect to $x$,we get $2y \frac{dy}{dx} = 8$,which implies $\frac{dy}{dx} = \frac{4}{y}$. The slope of the tangent $(m_t)$ at $(2, 4)$ is $\frac{4}{4} = 1$. The slope of the normal $(m_n)$ is $-\frac{1}{m_t} = -1$. The equation of the normal at $(x_1, y_1)$ with slope $m_n$ is given by $y - y_1 = m_n(x - x_1)$. Substituting the values,we get $y - 4 = -1(x - 2)$. $y - 4 = -x + 2$. $x + y = 6$.

Find the equation of the normal to the parabola $y^2 = 8x$ at the point $(2, 4)$.

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