Question

25) Find the area of the portion of the parabola y2 = 4ax cut off by the line y = ax
सरल रेखा y = ax द्वारा परवलय y = 4ax से काटे हुए क्षेत्र का क्षेत्रफल निकालें
OD (29)​

Answer 1

HELLO!

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Answer 2

The area of the portion of parabola cut by the line is $A=\dfrac{8}{3a}$

Step-by-step explanation:

Please find the attachment for figure and see common region.

Area of shaded region, $A=\int (y_{upper}-y_{lower})dx$

Upper curve of the common region is parabola,

• $y_{upper}=\sqrt{4ax}$

Lower curve of the common region is line,

• $y_{lower}=ax$

$A=\int_0^{4/a}(\sqrt{4ax}-ax)dx$

$A=(\dfrac{2}{3}\sqrt{4a}x^{3/2}-\dfrac{ax^2}{2})|_0^{4/a}$

$A=\dfrac{2}{3}\cdot 2\cdot\sqrt{a}\codt\dfrac{8}{a\sqrt{a}}-\dfrac{a}{2}\cdot\dfrac{16}{a^2}$

$A=\dfrac{32}{3a}-\dfrac{8}{a}$

$A=\dfrac{32-24}{3a}$

$A=\dfrac{8}{3a}$

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