26. Find the area enclosed between the parabola y2 = 4ax and the line y = mx

The equation of the parabola is y2=4a2 -----------(1)and that of line is y=mx ------(2)The Point of intersection of(1)and (2) is given by(mx)2=4ax⇒m2x2−4ax=0⇒x(m2x−4a)=0⇒x=0&x=4am2For, x=0,y2=4a×0→y=0 i.e, O(0,0)For, x=4am2,y2=4a×4am2→y=4am (in first quadrant)i.e, A(4am2,4am)Hence, the required area enclosed by the curve and the lines is a r e a ( D A C O ) = a r e a ( O C A B O ) − a r e a ( ? O A B )

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26. Find the area enclosed between the parabola y² = 4ax and the line y = mx

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Vishal Baghel | Contributor-Level 10

4 months ago

The equation of the parabola is $y^{2} = 4 a^{2}$ -----------(1)

and that of line is $y = m x$ ------(2)

The Point of intersection of(1)and (2) is given by

$\begin{array}{l} {(m x)}^{2} = 4 a x \\ \Rightarrow m^{2} x^{2} - 4 a x = 0 \\ \Rightarrow x (m^{2} x - 4 a) = 0 \\ \Rightarrow x = 0 & x = \frac{4 a}{m^{2}} \end{array}$

For, $x = 0, y^{2} = 4 a \times 0 \to y = 0$ i.e, O(0,0)

For, $x = \frac{4 a}{m^{2}}, y^{2} = 4 a \times \frac{4 a}{m^{2}} \to y = \frac{4 a}{m}$ (in first quadrant)

i.e, $A (\frac{4 a}{m^{2}}, \frac{4 a}{m})$

Hence, the required area enclosed by the curve and the lines is

$a r e a (D A C O) = a r e a (O C A B O) - a r e a (? O A B)$

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