The equation of the normal to the parabola x^ | vedclass

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(D) Given the parabola equation is $x^2 = 8y$. When $x = 4$,we substitute into the equation: $4^2 = 8y \Rightarrow 16 = 8y \Rightarrow y = 2$. So,the

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$x + y = 6$

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(D) Given the parabola equation is $x^2 = 8y$. When $x = 4$,we substitute into the equation: $4^2 = 8y \Rightarrow 16 = 8y \Rightarrow y = 2$. So,the point of contact is $(4, 2)$. Differentiating $x^2 = 8y$ with respect to $x$,we get $2x = 8 \frac{dy}{dx}$,which implies $\frac{dy}{dx} = \frac{2x}{8} = \frac{x}{4}$. The slope of the tangent at $x = 4$ is $m_t = \left. \frac{x}{4} \right|_{x=4} = 1$. The slope of the normal $m_n$ is given by $-\frac{1}{m_t} = -\frac{1}{1} = -1$. The equation of the normal at $(4, 2)$ is $y - y_1 = m_n(x - x_1)$. Substituting the values: $y - 2 = -1(x - 4)$. $y - 2 = -x + 4$. $x + y = 6$.

The equation of the normal to the parabola $x^2 = 8y$ at $x = 4$ is

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