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Difference between revisions of "1971 AHSME Problems/Problem 19"

(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)
 
Line 15: Line 15:
 
<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 = \frac{3}{4}.</math> The answer is <math>\textbf{(C)}.</math>
+
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 $y=mx+1$ intersects the ellipse $x^2+4y^2=1$ exactly once, then the value of $m^2$ is

$\textbf{(A) }\textstyle\frac{1}{2}\qquad \textbf{(B) }\frac{2}{3}\qquad \textbf{(C) }\frac{3}{4}\qquad \textbf{(D) }\frac{4}{5}\qquad \textbf{(E) }\frac{5}{6}$

Solution

Plug in $y=mx+1$ into the ellipse's equation to find the intersection points: \[x^2 + 4(mx+1)^2 = 1\] After simplifying, we have a quadratic in $x$: \[(4m^2 + 1)x^2 + 8mx +3 = 0\]

Because there is only one intersection point, then the quadratic has only one solution. This can only happen when the discriminant is 0. \[\Delta = b^2 - 4ac = (8m)^2 - (4)(1+4m^2)(3) = 0\]

Solving, we find $m^2 = \boxed{\textbf{(C) }\tfrac34}.$

-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. AMC logo.png

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