Try this beautiful Problem on Algebra based on Problem on Curve from AMC 10 A, 2018. You may use sequential hints to solve the problem.
Curve- AMC 10A, 2018- Problem 21
Which of the following describes the set of values of $a$ for which the curves $x^{2}+y^{2}=a^{2}$ and $y=x^{2}-a$ in the real $x y$ -plane intersect at
exactly 3 points?
- $a=\frac{1}{4}$
- $\frac{1}{4}<a<\frac{1}{2}$
- $a>\frac{1}{4}$
- $a=\frac{1}{2}$
- $a>\frac{1}{2}$
Key Concepts
Algebra
greatest integer
Suggested Book | Source | Answer
Suggested Reading
Pre College Mathematics
Source of the problem
AMC-10A, 2018 Problem-14
Check the answer here, but try the problem first
$a>\frac{1}{2}$
Try with Hints
First Hint
We have to find out the value of \(a\)
Given that $y=x^{2}-a$ . now if we Substitute this value in $x^{2}+y^{2}=a^{2}$ we will get a quadratic equation of $x$ and \(a\). if you solve this equation you will get the value of \(a\)
Now can you finish the problem?
Second Hint
After substituting we will get $x^{2}+\left(x^{2}-a\right)^{2}$=$a^{2} \Longrightarrow x^{2}+x^{4}-2 a x^{2}=0 \Longrightarrow x^{2}\left(x^{2}-(2 a-1)\right)=0$
therefore we can say that either \(x^2=0\Rightarrow x=0\) or \(x^2-(2a-1)=0\)
\(\Rightarrow x=\pm \sqrt {2a-1}\). Therefore
Now Can you finish the Problem?
Third Hint
Therefore \(\sqrt {2a-1} > 0\)
\(\Rightarrow a>\frac{1}{2}\)
Other useful links
- https://www.cheenta.com/surface-area-of-cube-amc-10a-2007-problem-21/
- https://www.youtube.com/watch?v=y-8Ru_qKDxk