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# Graph Coordinates | AMC 10A, 2015 | Question 12

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

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

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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

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

## Subscribe to Cheenta at Youtube

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