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Problem on Curve | AMC 10A, 2018 | Problem 21

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


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


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 \(\sqrt {2a-1} > 0\)

\(\Rightarrow a>\frac{1}{2}\)

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