## HOW TO FIND COMMON AREA OF TWO PARABOLAS ,AREA UNDER TWO PARABOLAS , AREA UNDER CURVES

## How to find common area of two parabolas , Area under two parabolas , Area of region bounded by two parabolas .

Let us consider two parabolas whose equations are given by

y

^{2 }= 4ax -------------- (1)

^{}x

^{2 }= 4ay ---------------- (2)

To check whether these parabolas intersect with each others or not And if they intersect then what is/are their point/s of intersection.

##
**How to find Points of Intersection**

To find coordinate of points of intersection ,we have to solve equation (1) and (2)
Consider eq (2)

x

⇒ y ^{2 }= 4ay^{ }= x

^{2 }/4a ---------------(3)

Putting the value of "y" in equation (1) ,we get

Area under Curves |

^{2 }/4a)

^{2 }= 4ax

⇒ x

^{4}/16a^{2 }= 4ax
⇒ x

⇒ x^{4}= 64xa^{3}^{4}- 64xa

^{3}= 0

Taking 'x' common

x(x

^{3}- 64a

^{3}) = 0

Either x = 0 or x

^{3}- 64a

^{3}= 0

⇒ x = 0 or (x)

^{3}- (4a)

^{3}= 0

⇒ x = 0 or (x)

^{3}- (4a)

^{3}= 0

⇒ x = 0 or (x-4a)

**[**(x)

^{2}+ (4a)

^{2}+ (x)(4a)

**]**= 0

⇒ x = 0 or (x-4a) = 0 or

**[**(x)

^{2}+ (4a)

^{2}+ (x)(4a)

**]**= 0

⇒ x = 0 or x = 4a or (x)

^{2}+ (4a)

^{2}+ (x)(4a) = 0

Since x

^{2}+ 4a.x + 16a

^{2}= 0 have

**no**real roots ,because its discriminant is negative, therefore this quadratic equation have complex roots. And these roots are rejected .

### To find values of y

Now putting both values of "x" in eq (3) i. e . y^{ }= x

^{2 }/4a ,we get

1st put x = 0

y

^{ }= 0 / 4a = 0 ⇒**when x = 0 then y = 0**
and put x = 4

y

^{ }= (4a)^{2 }/(4a)
y

^{ }= 4a ⇒**when x = 4a then y = 4a**
Hence two points of intersection of (1) and (2) O(0,0) and A (4a , 4a) .

Now draw two parabolas using their points of intersections as drawn in given picture.

## How to Find Required Area

Now to find the area enclosed between two Parabolas.

Required Area = shaded Area =Area OLAMO - Area ONAMO

**Watch this video to remove your doubts if any**

**My Previous Posts**

## Final words

^{}

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