Test: Ellipse- 1 - JEE MCQ

# Test: Ellipse- 1 - JEE MCQ

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## 10 Questions MCQ Test Mathematics (Maths) for JEE Main & Advanced - Test: Ellipse- 1

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Test: Ellipse- 1 - Question 1

### The length of the semi-latus-rectum of an ellipse is one third of its major axis, its eccentricity would be

Detailed Solution for Test: Ellipse- 1 - Question 1

Correct Answer :- a

Explanation : Semi latus rectum of ellipse = one half the last rectum

b2/a = 1/3*2a

b2 = 2a2/3

b = (2a/3)1/2...........(1)

So, b2/a = a(1-e2)

b2 = a2(1-e2)

Substituting from (1)

2a2/3 = a2(1-e2)

e2 = 1-2/3

e2 = 1/(3)1/2

Test: Ellipse- 1 - Question 2

### In the ellipse x2 + 3y2 = 9 the distance between the foci is

Detailed Solution for Test: Ellipse- 1 - Question 2

x2 + 3y2 = 9
⇒(x2)/9 + (y2)/3=1
⇒a2=9, b2=3
⇒e=[1−b2/a2]1/2
=(2/3)1/2
Therefore, distance between foci is =2ae = 2 × 3 × (2/3)1/2
=2(6)1/2

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Test: Ellipse- 1 - Question 3

### The eccentricity of the ellipse 9x2 + 5y2 – 30y = 0 is:

Detailed Solution for Test: Ellipse- 1 - Question 3

9x2 + 5y2 - 30y = 0
9x2 + 5(y−3)2 = 45
We can write it as : [(x-0)2]/5 + [(y-3)2]/9 = 1
Compare it with x2/a2 + y2/b2 = 1
e = [(b2 - a2)/b2]½
e = [(9-5)/9]1/2
e = (4/9)½
e = ⅔

Test: Ellipse- 1 - Question 4

The centre of the ellipse  is:

Detailed Solution for Test: Ellipse- 1 - Question 4

Centre of the ellipse is the intersection point of
x+y−1=0.........(1)
x−y=0............(2)
Substituting x from equation 2 in equation 1 two equations, we get,
2y=2,   y=1
Replacing, we get x=1
⇒(1,1) is the centre

Test: Ellipse- 1 - Question 5

= 1, the length of the major axis is

Detailed Solution for Test: Ellipse- 1 - Question 5

Test: Ellipse- 1 - Question 6

The eccentricity of an ellipse, with its centre at the origin, is 1/2. If one of the directrix is x = 4, then the equation of the ellipse is:

Detailed Solution for Test: Ellipse- 1 - Question 6

Let the equation of ellipse (x2)/(a2)+(y2)/(b2)=1
Here a > ba > b because the directrix is parallel to y axis.
b2=a2(1−e2)
Given e= 1/2
⇒b2 = (3/4)a2
But a/e=4
⇒a=2
Putting a=2 we get b= (3)1/2
​Required ellipse is (x2)/4+(y2)/3=1
⇒3x2+4y2=12

Test: Ellipse- 1 - Question 7

The eccentricity of an ellipse whose latus rectum is equal to distance between foci is:

Detailed Solution for Test: Ellipse- 1 - Question 7

Distance between the foci of an ellipse = length of latus rectum
i.e.  (2b2)/a=2ae
e=b2/a2
But e=[1−b2/a2]1/2
Then e=(1−e)1/2
Squaring both sides, we get
e+e−1=0
e=−1 ± (1 + 4)1/2]/2
(∵ Eccentricity cannot be negative)
e=[(5)1/2 − 1]/2

Test: Ellipse- 1 - Question 8

The foci of the ellipse 25 (x + 1)2 + 9(y + 2)2 = 225 are at:

Detailed Solution for Test: Ellipse- 1 - Question 8

Here equation of ellipse is 25(x + 1)2 + 9(y + 2)2 = 225
0r (x + 1)2/9 + (y + 2)2/25 = 1
Centre of the ellipse is (–1,–2)
a2 = 9, b2  = 25
a = 3, b = 5
e = (1-a2/b2)1/2
e = (1-9/25)1/2
e = +-4/5
be = +-4
Foci : (-1,-6)(-1,2)

Test: Ellipse- 1 - Question 9

The equation of the ellipse whose one focus is at (4, 0) and whose eccentricity is  4/5 is:

Detailed Solution for Test: Ellipse- 1 - Question 9

focus lies on x axis
So, the equation  of ellipse is x2/a2 + y2/b2 = 1
Co-ordinate of focus(+-ae, 0)
ae = 4
e = ⅘
a = 4/e  => 4/(⅘)
a = 5
(a)2 = 25
b2 = a2(1-e2)
= 25(1-16/25)
b2 = 9
Required equation : x2/(5)2 + y2/(3)2 = 1

Test: Ellipse- 1 - Question 10

The radius of the circle given by 2x2 + 2y2 – x = 0 is

Detailed Solution for Test: Ellipse- 1 - Question 10

2x² + 2y² - x = 0 .
==> 2 ( x² + y² - x/2 ) = 0 .
==> 2/2 ( x² + y² - x/2 ) = 0/2 .
==> x² - x/2 + y² = 0 .
==> ( x² - x/2 + (1/4)² ) + y² = (1/4)² .
==> ( x - 1/4 )² + ( y - 0 )² = (1/4)²
Centre (-¼, 0)    radius(¼)

## Mathematics (Maths) for JEE Main & Advanced

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## Mathematics (Maths) for JEE Main & Advanced

209 videos|443 docs|143 tests