Two tangent lines l_{1} and l_{2} are drawn f | vedclass

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(D) The equation of the parabola is $y^{2} = -\frac{1}{2}x$. Comparing with $y^{2} = 4ax$,we get $4a = -\frac{1}{2}$,so $a = -\frac{1}{8}$. The equati

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$9$

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(D) The equation of the parabola is $y^{2} = -\frac{1}{2}x$. Comparing with $y^{2} = 4ax$,we get $4a = -\frac{1}{2}$,so $a = -\frac{1}{8}$. The equation of a tangent to the parabola with slope $m$ is $y = mx + \frac{a}{m} = mx - \frac{1}{8m}$. Since the tangent passes through $(2,0)$,we have $0 = 2m - \frac{1}{8m}$,which implies $2m = \frac{1}{8m}$,so $m^{2} = \frac{1}{16}$,giving $m = \pm \frac{1}{4}$. The equations of the tangents are $y = \frac{1}{4}(x-2)$ and $y = -\frac{1}{4}(x-2)$,which simplify to $x - 4y - 2 = 0$ and $x + 4y - 2 = 0$. These lines are also tangent to the circle $(x-5)^{2} + y^{2} = r$. The distance from the center $(5,0)$ to the line $x \pm 4y - 2 = 0$ must equal the radius $\sqrt{r}$. Using the distance formula $d = \frac{|ax_{0} + by_{0} + c|}{\sqrt{a^{2} + b^{2}}}$,we get $\sqrt{r} = \frac{|5 - 0 - 2|}{\sqrt{1^{2} + (-4)^{2}}} = \frac{3}{\sqrt{17}}$. Squaring both sides,$r = \frac{9}{17}$. Therefore,$17r = 9$.

Two tangent lines $l_{1}$ and $l_{2}$ are drawn from the point $(2,0)$ to the parabola $2y^{2} = -x$. If the lines $l_{1}$ and $l_{2}$ are also tangent to the circle $(x-5)^{2} + y^{2} = r$,then $17r$ is equal to.

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