Area bounded by the ellipse 2 x 2-3 y 2-1 isA- -6B- 6 -C- -6-D- 1-6

The correct option is A π√(6)Equation of the ellipse can be written as x^21/2+y^21/3=1 ⇒ a=1√(2) and b=1√(3) Area of ellipse is given by A=π a b ∴ A=π√(6) ...

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Standard XII

Mathematics

Pair of Lines

Area bounded ...

Question

Area bounded by the ellipse $2 x^{2} + 3 y^{2} = 1$ is

\frac{π}{6}

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\frac{π}{\sqrt{6}}

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6 π

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\sqrt{6} π

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Solution

The correct option is B $\frac{π}{\sqrt{6}}$
Equation of the ellipse can be written as
$\frac{x^{2}}{1 / 2} + \frac{y^{2}}{1 / 3} = 1$
$\Rightarrow a = \frac{1}{\sqrt{2}}$ and $b = \frac{1}{\sqrt{3}}$
Area of ellipse is given by
$A = π a b$
$∴ A = \frac{π}{\sqrt{6}}$

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Similar questions

Q. Assertion :The area of the ellipse

2 x^{2} + 3 y^{2} = 6

will be more than the area bounded by

2 | x | + 3 | y | \leq 6.

Reason: The length of the major axis of the ellipse

2 x^{2} + 3 y^{2} = 6

is less than the distance between the points of

2 | x | + 3 | y | \leq 6

x -

axis.

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is _____

The area (in square units) bounded by the curves $x = - 2 y^{2} a n d x = 1 - 3 y^{2}$ is

Q. The line

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is tangent to the ellipse

2 x^{2} + 3 y^{2} = 1.

Then

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Q. Match the following:

List-I	List-II
1. Area bounded by $y = \| x \|$ and $y = 2$ is	a. 4
2. Area bounded by $\frac{\| x \|}{a} + \frac{\| y \|}{b} = 1$ , when $a, b > 0$ is	b. $\frac{(π - 2) a b}{4}$
3. Area between the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ and the chord $\frac{x}{a} + \frac{y}{b} = 1$	c. 1
4. Area bounded by $y = [x]$ , the $x$ -axis and $x = 1, x = 2$ is	d. 2ab

The correct match is

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Area bounded by the ellipse 2x2+3y2=1 is

The correct option is B π√6Equation of the ellipse can be written as x21/2+y21/3=1 ⇒a=1√2 and b=1√3 Area of ellipse is given by A=πab ∴A=π√6

Area bounded by the ellipse $2 x^{2} + 3 y^{2} = 1$ is

The correct option is B $\frac{π}{\sqrt{6}}$
Equation of the ellipse can be written as
$\frac{x^{2}}{1 / 2} + \frac{y^{2}}{1 / 3} = 1$
$\Rightarrow a = \frac{1}{\sqrt{2}}$ and $b = \frac{1}{\sqrt{3}}$
Area of ellipse is given by
$A = π a b$
$∴ A = \frac{π}{\sqrt{6}}$