MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Summation by Sigma Method

The area betw...

Question

The area between the parabolas $y^{2} = 4 a (x + a)$ and $y^{2} = - 4 a (x - a)$ in sq. units is

A

\frac{4 a^{2}}{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

\frac{8 a^{2}}{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C

\frac{12 a^{2}}{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D

\frac{16 a^{2}}{3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is C $\frac{16 a^{2}}{3}$
$y^{2} = 4 a (x + a)$ and $y^{2} = - 4 a (x - a)$
$4 0 \int - a \sqrt{4 a (x + a)} d x = 4 \times 2 \sqrt{a} \times \frac{2}{3} (x + a)^{3 / 2} |_{- a}^{0}$
$= \frac{16}{3} \sqrt{a} (a^{3 / 2})$
$= \frac{16 a^{2}}{3} s q u n i t s .$

Suggest Corrections

thumbs-up

0

Similar questions

Q. Assertion(A): The area bounded by

y^{2} = 4 x

x^{2} = 4 y

\frac{16}{3}

sq. units.

Reason(R): The area bounded by

y^{2} = 4 a x

x^{2} = 4 a y

\frac{16 a^{2}}{3}

Q. Prove that the area enclosed between two parabolas

y^{2} = 4 a x

x^{2} = 4 a y

\frac{16 a^{2}}{3}

Q. The area enclosed between the parabola

y^{2} = 4 a x

x = a, x = 9 a

in sq. units is:

Q. The area bounded by

y^{2} = 4 a x

y = m x

\frac{a^{2}}{3}

m

Q. Match the following:

List-I	List-II
1. Area bounded by $y^{2} = 4 a x$ and $x^{2} = 4 a y$ is	a. $\frac{8 a^{2}}{3}$
2. Area bounded by $y^{2} = 4 a x$ and its latusrectum is	b. $\frac{a^{2}}{6}$
3. Area bounded by $x = 2 c o s θ, y = 2 s i n θ$ is	c. $\frac{16 a^{2}}{3}$
4. Area bounded by $\sqrt{x} + \sqrt{y} = \sqrt{a}$ is	d. $2 π$

The correct match is

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Summation by Sigma Method

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Summation by Sigma Method

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon