Test: Ellipse (21 July) - JEE MCQ

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

Equation of ellipse : x²/9 + y²/1= 1
Co−ordinates of point P is : (acosθ , bsinθ)
Equation of normal at point (x1, y1) is :
a²x/x1 − b²y/y1 = (ae)²
Equation of normal at point P is:
ax/cosθ− by/sinθ= (ae)²−−−−−(1)
Equation of diameter at Q is : y = (−b/a)tanθ−−−−−(2)
Point of intersection of equation 1 and 2 is :
ax/cosθ + (b²/acosθ) = (ae)²
Or , x = ae²cosθ − b²/a²
And y = − be²sinθ + b³/a³ (from 2)
∴ Point R(ae²cosθ − b²/a² , − be²sinθ + b³/a³)
Let mid point of PR is (h,k)
h =[acosθ + ae²cosθ − (b²/a²)]/2
Or , 2h + b²/a²= acosθ + ae²cosθ
Or , cosθ= [2h + (b²/a²)/(a + ae²)]
k = [bsinθ − be²sinθ + (b³/a³/2)]/2
Or , sinθ = [2k − (b³/a³)/(b − be²)]
On squaring adding we get :
[2h + (b²/a²)]/(a + ae²)² + [2k − (b³/a³)] / (b − be²)²= 1
Which is an equation of ellipse.

The eccentricity for ellipse is always less than 1. The eccentricity is always 1 for any parabola. The eccentricity is always 0 for a circle. The eccentricity for a hyperbola is always greater than 1.

Given,  x²/5 + y²/3= 1
We know S₁F₁ × S₂F₂ = b²
∴ S₁F₁ × S₂F₂ = 3

Clearly, they subtend right angle at centre.

Ellipse passing through O (0, 0) and having foci P (3, 3) and Q (-4, 4) ,
Then 

is an ellipse, whose focus is (2, -3) , directrix 3x - 4y + 7 = 0 and eccentricity is 1/2.
Length offrom focus to directrix is 



So length of major axis is 20/3

Any point on the ellipse is (2 cosθ, √3 sinθ). The focus on the positive x-axis is (1,0). 

Given that (2cosθ-1)² + 3sin² θ = 25/16
⇒ cos θ = 3/4

Let point Q be (h,k) , where k  < 0
Given that k = SP = a + ex₁ , where P (x₁,y₁) lies on the ellipse

Let  be the tangent
It is passing through(-2,0)