Test: Ellipse- 1 - JEE MCQ

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

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

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

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

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

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
e² +e−1=0
e=−1 ± (1 + 4)^1/2]/2
(∵ Eccentricity cannot be negative)
e=[(5)^1/2 − 1]/2

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)

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

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)