Algebra> Prove that the curves y2 = 4x and x2 = 4y...

Prove that the curves y2 = 4x and x2 = 4y divide the area of the square bounded by x = 0, x = 4, y = 4 and y = 0 into three equal parts.?

sudhanshu , 10 Years ago

Grade 12

1 Answers

Jitender Singh

Last Activity: 10 Years ago

Ans:

Area of square ‘A’:

$A = 4 \times 4 =16$

Area b/w curves ‘a’:

$a = \int_{0}^{4}(2\sqrt{x} - \frac{x^{2}}{4})dx$

$a = (\frac{4x^{3/2}}{3}-\frac{x^{3}}{12})_{0}^{4} = \frac{32}{3}-\frac{16}{3} = \frac{16}{3}$

Area of rest two parts:

$A -a = \frac{32}{3}$

which is equally divided into 16/3 & 16/3.

Thanks & Regards

Jitender Singh

IIT Delhi

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Algebra> Prove that the curves y2 = 4x and x2 = 4y...

Prove that the curves y2 = 4x and x2 = 4y divide the area of the square bounded by x = 0, x = 4, y = 4 and y = 0 into three equal parts.?

sudhanshu , 10 Years ago

Jitender Singh

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LIVE ONLINE CLASSES

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