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HMMT 二月 1998 · CALC 赛 · 第 8 题

HMMT February 1998 — CALC Round — Problem 8

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

Question Eight . [6 points]
Find the slopes of all lines passing through the origin and tangent to
2 3
the curve y = x + 39 x − 35.

解析

英文解析

  1. Problem: Find the slopes of all lines passing through the origin and tangent to the curve
    2 3
    y = x + 39 x − 35.
    Solution: Any line passing throug the origin has equation y = mx , where m is the slope of the line. If adyline is tangent to the given curve, then at the point of tangency, ( x, y ), = m .
    2 dxdy dy
    2 3 x +39
    First, we calculate of the curve: 2 ydy = 3 x dx + 39 dx ⇒ = . Substituting mx for y , we getdx dx 2 ythe following system of equations:
    2 2 3
    m x = x + 39 x − 35
    3 x + 392
    m =
    2 mx
    Solving for x yields the equation x − 39 x + 70 = 0 ⇒ ( x − 2)( x + 7)( x − 5) = 0 ⇒ x = 2 or x = − 7 or 3
    x = 5. These solutions indicate the x -coordinate of the points at which the desired lines are tangent to the

    curve. Solving for the slopes of these lines, we get m = ± for x = 2, no real solutions for x = − 7, and 51
    √ √ √2
    285 51 285
    m = ± for x = 5. Thus m = ± , ± .
    5 2 5

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