An equation for the line that passes through | vedclass

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(D) Given the parabola $y = \frac{x^2}{4} - 2$,we have $4y = x^2 - 8$.Differentiating with respect to $x$,we get $4 \frac{dy}{dx} = 2x$,so $\frac{dy}{

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$x + 2y = 8$

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(D) Given the parabola $y = \frac{x^2}{4} - 2$,we have $4y = x^2 - 8$.Differentiating with respect to $x$,we get $4 \frac{dy}{dx} = 2x$,so $\frac{dy}{dx} = \frac{x}{2}$.Let the point of contact be $P(x_1, y_1)$. The slope of the tangent at $P$ is $m_t = \frac{x_1}{2}$.The slope of the normal at $P$ is $m_n = -\frac{1}{m_t} = -\frac{2}{x_1}$.The line passes through $A(10, -1)$ and $P(x_1, y_1)$,so its slope is $\frac{y_1 - (-1)}{x_1 - 10} = \frac{y_1 + 1}{x_1 - 10}$.Since this line is the normal,$\frac{y_1 + 1}{x_1 - 10} = -\frac{2}{x_1}$.$x_1(y_1 + 1) = -2(x_1 - 10)$ $\Rightarrow x_1y_1 + x_1 = -2x_1 + 20$ $\Rightarrow x_1y_1 + 3x_1 = 20$.Substituting $y_1 = \frac{x_1^2}{4} - 2$,we get $x_1(\frac{x_1^2}{4} - 2) + 3x_1 = 20$.$\frac{x_1^3}{4} - 2x_1 + 3x_1 = 20$ $\Rightarrow \frac{x_1^3}{4} + x_1 = 20$ $\Rightarrow x_1^3 + 4x_1 - 80 = 0$.By inspection,$x_1 = 4$ is a root: $64 + 16 - 80 = 0$.For $x_1 = 4$,$y_1 = \frac{4^2}{4} - 2 = 4 - 2 = 2$. So $P = (4, 2)$.The slope of the normal is $m_n = -\frac{2}{4} = -\frac{1}{2}$.The equation of the line passing through $(10, -1)$ with slope $-\frac{1}{2}$ is $y - (-1) = -\frac{1}{2}(x - 10)$.$2(y + 1) = -(x - 10)$ $\Rightarrow 2y + 2 = -x + 10$ $\Rightarrow x + 2y = 8$.

An equation for the line that passes through $(10, -1)$ and is perpendicular to $y = \frac{x^2}{4} - 2$ is

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