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...")
 
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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.
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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>
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Solving, we find <math>m^2 = \boxed{\textbf{(C) }\tfrac34}.</math>
  
 
-edited by coolmath34
 
-edited by coolmath34
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== See Also ==
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{{AHSME 35p box|year=1971|num-b=18|num-a=20}}
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{{MAA Notice}}

Latest revision as of 08:44, 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 (ProblemsAnswer KeyResources)
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

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