Olympiad Maths Prep

Track / Stage 6 / 314 of 400 #1314 of 2000

Problem 1314

National olympiad, first round
Algebra Difficulty 6.6 Find the answer

A line through the origin passes through the curve whose equation is 5y=2x29x+105y=2x^2-9x+10 at two points whose xx-coordinates add up to 77.77. Find the slope of the line.

Official solution

1. Let the equation of the line through the origin be y=kx y = kx , where k k is the slope of the line.
2. Substitute y=kx y = kx into the given equation of the parabola 5y=2x29x+10 5y = 2x^2 - 9x + 10 :
5(kx)=2x29x+10 5(kx) = 2x^2 - 9x + 10
Simplify this to:
5kx=2x29x+10 5kx = 2x^2 - 9x + 10
3. Rearrange the equation to form a standard quadratic equation:
2x2(9+5k)x+10=0 2x^2 - (9 + 5k)x + 10 = 0
4. By Vieta's formulas, the sum of the roots of the quadratic equation ax2+bx+c=0 ax^2 + bx + c = 0 is given by ba -\frac{b}{a} . Here, a=2 a = 2 and b=(9+5k) b = -(9 + 5k) , so the sum of the roots is:
9+5k2 \frac{9 + 5k}{2}
5. We are given that the sum of the x x -coordinates of the points where the line intersects the parabola is 77. Therefore, we set up the equation:
9+5k2=77 \frac{9 + 5k}{2} = 77
6. Solve for k k :
9+5k=154 9 + 5k = 154
5k=145 5k = 145
k=29 k = 29

The final answer is 29 \boxed{29} .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.