Try this beautiful Problem on Graph Coordinates from coordinate geometry from AMC 10A, 2015.

## Graph Coordinates – AMC-10A, 2015- Problem 12

Points $(\sqrt{\pi}, a)$ and $(\sqrt{\pi}, b)$ are distinct points on the graph of $y^{2}+x^{4}=2 x^{2} y+1 .$ What is $|a-b| ?$

,

- $0$
- $1$
- $2$
- $3$
- \(4\)

**Key Concepts**

Co-ordinate geometry

graph

Distance Formula

## Check the Answer

Answer: $2$

AMC-10A (2015) Problem 12

Pre College Mathematics

## Try with Hints

The given points are $(\sqrt{\pi}, a)$ and $(\sqrt{\pi}, b)$ which are satisfying the equation $y^{2}+x^{4}=2 x^{2} y+1$.

So we can write $y^{2}+\sqrt{\pi}^{4}=2 \sqrt{\pi}^{2} y+1$

Can you now finish the problem ……….

Therefore

$y^{2}+\pi^{2}=2 \pi y+1$

$y^{2}-2 \pi y+\pi^{2}=1$

$(y-\pi)^{2}=1$

$y-\pi=\pm 1$

$y=\pi+1$

$y=\pi-1$

can you finish the problem……..

There are only two solutions to the equation, so one of them is the value of $a$ and the other is $b$. our requirement is $|a-b|$ so between a and b which is greater is not importent…………

So, $|(\pi+1)-(\pi-1)|=2$

## Other useful links

- https://www.cheenta.com/surface-area-of-cube-amc-10a-2007-problem-21/
- https://www.youtube.com/watch?v=h_x9kS-J1XY&t=41s