1956 AHSME Problems/Problem 21

Problem 21

If each of two intersecting lines intersects a hyperbola and neither line is tangent to the hyperbola, then the possible number of points of intersection with the hyperbola is:

$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 2\text{ or }3\qquad \textbf{(C)}\ 2\text{ or }4\qquad \textbf{(D)}\ 3\text{ or }4\qquad \textbf{(E)}\ 2,3,\text{ or }4$

Solution

Consider the hyperbola $x^2-y^2=1$. It is possible for the two intersecting lines to intersect the hyperbola at 2 points if one of them has a slope of 1 and only intersects one part of the hyperbola and the other line doesn't intersect the hyperbola at all (Ex. $y=x+3,\, x=0.$). If the second line is instead $x=4$, it intersects the hyperbola twice, so the lines can intersect the hyperbola 3 times. Finally, if both lines intersect the hyperbola twice, such as $y=2x-4$ and $y=3x-6$, the lines can intersect the hyperbola 4 times. So the answer is $\textbf{(E)}\ 2,3,\text{ or } 4$

See Also

1956 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 22
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png