MyQuestionIcon

1

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

Surface Area of Combined Figures

The ratio of ...

Question

The ratio of the area enclosed by the curve $a x^{2} + b y^{2} = a b; a, b > 0$ divided by the straight line $y = m x, m \neq 0,$ is

A

a : b

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

B

a b : m^{2}

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

C

1 : 1

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

D

a m^{2} : b

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

Open in App

Solution

The correct option is C $1 : 1$
Curve shown in the figure :
We can see that, the enclosed area by ellipse is divided into two equal halves.
So, ratio of area is $1 : 1$

Suggest Corrections

thumbs-up

0

Similar questions

Q. If the area bounded by the curve

4 x^{2} - 8 x + y^{2} - 2 y + 1 = 0

is divided into two parts by straight line

2 x + y = 5,

then the ratio of smaller to the larger area is

y = m x

bisects the area enclosed by the curve

y = 1 + 4 x - x^{2}

x = 0, x = \frac{3}{2}

y = 0

. Then the value of

m

y = m x

bisects the area enclosed by the lines

x = 0,

y = 0,

x = \frac{3}{2}

y = 1 + 4 x - x^{2},

12 m

Q. Find the area enclosed by the curve

y = - x^{2}

and the straight line x + y + 2 = 0. [NCERT EXEMPLAR]

The ratio in which the area enclosed by the curve $y = cos x (0 \leq 0 \leq \frac{π}{2})$ in the first quadrant is divided by the curve $y = sin x$ , is:

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Combination of Solids

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Surface Area of Combined Figures

Standard X Mathematics

Join BYJU'S Learning Program

CrossIcon