Difference between revisions of "1971 AHSME Problems/Problem 19"
Coolmath34 (talk | contribs) (Created page with "== Problem == If the line <math>y=mx+1</math> intersects the ellipse <math>x^2+4y^2=1</math> exactly once, then the value of <math>m^2</math> is <math>\textbf{(A) }\textstyl...") |
m (see also, boxed answer, added link) |
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<cmath>(4m^2 + 1)x^2 + 8mx +3 = 0</cmath> | <cmath>(4m^2 + 1)x^2 + 8mx +3 = 0</cmath> | ||
− | Because there is only one intersection point, then the quadratic has only one solution. This can only happen when the discriminant is 0. | + | Because there is only one intersection point, then the quadratic has only one solution. This can only happen when the [[discriminant]] is 0. |
<cmath>\Delta = b^2 - 4ac = (8m)^2 - (4)(1+4m^2)(3) = 0</cmath> | <cmath>\Delta = b^2 - 4ac = (8m)^2 - (4)(1+4m^2)(3) = 0</cmath> | ||
− | Solving, we find <math>m^2 = \ | + | Solving, we find <math>m^2 = \boxed{\textbf{(C) }\tfrac34}.</math> |
-edited by coolmath34 | -edited by coolmath34 | ||
+ | |||
+ | == See Also == | ||
+ | {{AHSME 35p box|year=1971|num-b=18|num-a=20}} | ||
+ | {{MAA Notice}} |
Latest revision as of 09:44, 5 August 2024
Problem
If the line intersects the ellipse
exactly once, then the value of
is
Solution
Plug in into the ellipse's equation to find the intersection points:
After simplifying, we have a quadratic in
:
Because there is only one intersection point, then the quadratic has only one solution. This can only happen when the discriminant is 0.
Solving, we find
-edited by coolmath34
See Also
1971 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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 • 31 • 32 • 33 • 34 • 35 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.