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สมัครสมาชิก | คู่มือการใช้ | รายชื่อสมาชิก | ปฏิทิน | ข้อความวันนี้ | ค้นหา |
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เครื่องมือของหัวข้อ | ค้นหาในหัวข้อนี้ |
#1
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ทุนญี่ปุ่น 2019B
Consider the circle $x^{2}+y^{2}=r^{2} (r>0)$ and the line $y=m(x-5)+3$. The line intersects the circle if and only if $0\leqslant m\leqslant ?$. Also $r=?$.
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#2
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อ้างอิง:
Now, we have to calculate the possible range of $m$. Since $r=3$ has been established, we let $A=(0,3)$ be a point such that $PA$ is tahgnet to the circle $C$ given by the equation $x^2+y^2=r^2=3^2=9$. Suppose that $B\neq A$ is another point on the circle $C$ such that $PB$ is tangent to $C$. Suppose that $B=(u,v)$ for some coordinates $u$ and $v$. Then, we know that $AB\perp OP$, where $O$ is the origin, which is the center of $C$. Since the equation of the line $OP$ is $y=\dfrac{3}{5}\,x$, the equation of the line $AB$ must be $$y=-\frac{5}{3}\,x+3\,.$$ Therefore, $u^2+v^2=9$ because $B$ lies on $C$, as well as $v=-\frac{5}{3}\,u+3$, because $B$ lies on $AB$. This gives $$u^2+\left(-\frac{5}{3}\,u+3\right)^2=9\,.$$ Hence, $$u^2+\frac{25}{9}\,u^2-10\,u+9=9\,.$$ Consequently, $$\frac{34}{9}\,u^2-10\,u=0\,.$$ Because $u\neq 0$ (otherwise $v=3$, so $B=A$, which contradicts the assumption that $B\neq A$), we get $$u=\frac{45}{17}\,.$$ Therefore, $$v=-\frac{5}{3}\,u+3=-\frac{75}{17}+3=-\frac{24}{17}\,.$$ The maximum value of $m$ is thus $$\frac{3-v}{5-u}=\frac{3+\frac{24}{17}}{5-\frac{45}{17}}=\frac{51+24}{85-45}=\frac{75}{40}=\frac{15}{8}\,.$$
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Потом доказывай, что ты не верблюд. 28 กรกฎาคม 2020 21:18 : ข้อความนี้ถูกแก้ไขแล้ว 2 ครั้ง, ครั้งล่าสุดโดยคุณ Anton |
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