The area between the parabola $y^{2} = 4 x$ , normal at one end of latus rectum and $x - a x i s$ in sq. units is:

\frac{1}{3}

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\frac{2}{3}

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\frac{10}{3}

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\frac{4}{3}

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Solution

The correct option is D $\frac{4}{3}$
R.E.F image
$y^{2} = 4 x$
area = ?
$2 \int \sqrt{x} = 2 \times 2 (x)^{3 / 2}$
$= \frac{4}{3} (1)$

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Similar questions

The area bounded by the normal at $(1, 2)$ to the parabola $y^{2} = 4 x$ , x-axis and the curve is given by

y^{2} = 4 (x - 1)

x^{2} + 4 (y - 3) = 0

A

x^{2} - 2 y^{2} - 2 \sqrt{5} x - 4 \sqrt{2} y - 3 = 0

B

C

A, B

C

A B C

y^{2} + 4 x + 4 y - 3 = 0

y = 4 x^{2} - 4 x + 3

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