Ellipse MCQ Quiz - Objective Question with Answer for Ellipse - Download Free PDF

Calculation 

 = 1 foci are (ae, 0) and (–ae, 0) 

 = 1  foci are (Ae', 0) and (–Ae', 0) 

⇒ 2ae = ⇒ ae =

and 2Ae' = ⇒ Ae' = 

⇒ ae = Ae' ⇒ 

⇒  ⇒  a = 3A

Now a – A = 2 ⇒ a – – 2 ⇒ a = 3 and A = 1 

Ae =   ⇒ e =  and e' = 

b² = a²(1 – e²)

b² = 6

and B² = A²((e')² – 1) = (2) ⇒ B² = 2

sum of LR =  = 8

Hence option 3 is correct

Explanation:

Given:

4x2 + 9y2 = 1 

Here a = 1/2 and b = 1/3

Now 

e = 

foci = (±ae, 0)

= ( ±

= (± )

Thus, P(x, y) is any point on the ellipse

⇒ PQ + PR = 2a = 2× 1/2 = 1

∴ Option (b) is correct.

Equation of chord T = S₁ 

⇒ 

⇒ 40x + 18y = 109

⇒ α = 40, β = 18

⇒ α + β = 58

Concept:

If the equation of an ellipse is, then the coordinates of foci will be (ae,0) and (-ae,0)

where e is the eccentricity of the ellipse

and b2 = a2(1 - e²)

Calculation:

Let the equation of ellipse be 

then the coordinates of foci will be (ae,0) and (-ae,0), where e is the eccentricity

Given foci at (-4,0) and (4,0),

⇒ ae = 4 or e = 4/a

and b2 = a2(1 - e2) = 

⇒ b² = a² - 16 ___(1)

Since the ellipse  passing through (3, ) 

⇒ 

Putting the value from (1),

⇒ 

⇒ 18(a² - 16) + 10a² = a²(a² - 16)

⇒ a⁴ - 44a² + 288 = 0

⇒ (a² - 36)(a² - 8) = 0

⇒ a² = 36, 8

From (1),

⇒ b² = 20, -8

Since b can't be negative

So, a² = 36 and b² = 20

So the eccentricity = e = 4/a = 4/6

⇒ e = 

∴ The correct option is (5).

Concept:

The equation of ellipse with centre (0,0) is of the form, 

For an ellipse  with y-axis as major axis, the eccentricity = 

Calculation:

Given: The centre of an ellipse is at (0, 0), major axis is on the y-axis and the ellipse passes through (3, 2) and (1, 6).

Let the equation of the ellipse be 

(3,2) and (1,6) lies on the ellipse

⇒  and 

⇒  __(i)

and  __(ii)

Subtracting (i) from (ii),

⇒ 

 ⇒ 32a²​ = 8b²

⇒ 4a² = b²

Put in (i) 

⇒ 

⇒ a² = 10

⇒ b² = 4 × 10 = 40 

So, the equation of the ellipse will be 

Now for the ellipse , the eccentricity of the ellipse is given by,

⇒ 

⇒ Eccentricity = 

∴ The correct option is (1).

Concept:

Equation of ellipse: ​

Eccentricity (e) = 

Where, vertices = (± a, 0) and focus = (± ae, 0)

Calculation:

Here, vertices of ellipse (± 5, 0) and foci (±4, 0)

So, a = ±5 ⇒  and

ae = 4 ⇒ e = 4/5

Now, 4/5 = 

∴ Equation of ellipse = 

Hence, option (1) is correct. 

Concept: 

Standard equation of ellipse ,  

Length of latus rectum , L.R =  , if  b > a

Calculation: 

 ,

On comparing with standard equation , a = 5 and b = 7 

We know that , Length of latus rectum =   

⇒ L.R =  =   . 

The correct option is 2. 

Concept:

Tangent to an Ellipse:

The equation of the tangent to the ellipse , at a point (x₁, y₁), is given by: .

Calculation:

At x = 3, we will have:

⇒ 

⇒ y = ± 

From the above formula, we can say that the required equation of the tangent to the ellipse  at  and  are (respectively):

⇒

⇒ 3x + 5y = 25

OR

⇒ 

⇒ 3x - 5y = 25

Concept:

The standard equation of an ellipse:

, a > b

Where 2a and 2b are the length of the major axis and minor axis respectively and center (0, 0)

The eccentricity = 

Length of latus rectum = 

Distance from center to focus = 

Calculation:

Given ellipse 

a² = 100 ⇒ a = 10

and b² = 64 ⇒ b = 8

The eccentricity (e)

⇒ e = 

⇒ e = 

⇒ e = 

⇒ e = 0.6

Now distance between foci = 2ae

= 2 × 10 × 0.6

∴ Distance between foci = 12

Concept:

Calculation:

25x² + 16y² = 400

Comparing, with standard equation: a = 4 ; b = 5

Since ( a

Concept:

The length of the latus rectum of the ellipse  is equal to . (a

Calculation:

Writing the equation of the ellipse in the standard form , we get:

∴ a = 2 and b = 2√3.

Here a

Length of the latus rectum =  =  = .

Concept:

The general equation of the ellipse is:

Here, coordinates of foci are (±ae, 0).

Also, we have b² = a²(1 - e²), where e is the eccentricity.

Calculation:

Since the coordinates of the foci are (±3, 0).

⇒ ae = 3

⇒ a × (1/3) = 3      (∵ e = 1/3)

⇒ a = 9

Now, b2 = a2(1 - e2)

⇒ b² = 72

On putting the value of a² and b² in the general equation of an ellipse, we get

Hence, the equation of the ellipse is .

Concept:

1. Equation of circle x2 + y2 = r2

2. Equation of an ellipse 

3. Equation of Parabola y2 = 4ax, x2 = 4ay

4. Equation of hyperbola 

5.If a2 = b2 then hyperbola is called rectangular hyperbola and x2 − y2 = a2 is the general form of a rectangular hyperbola

Calculation:

Given:

x = 3(cost + sint)

y = 4(cost - sint) 

 ----(1)

 ----(2)

Adding r=equation 1 and 2;

Hence, the given curve represents an ellipse.

Concept:

Standard equation of an ellipse:  (a > b)

• Coordinates of foci = (± ae, 0)
• Eccentricity (e) =  ⇔ a2e2 = a2 – b2
• Length of Latus rectum = 

Calculation:

Given: 

Compare with the standard equation of an ellipse: 

So, a² = 100 and b² = 75

∴ a = 10

Length of latus rectum =  = 

Concept:

The standard equation of an ellipse is given by:  

The sum of the focal distance of any point on an ellipse is constant and equal to the length of the major axis of the ellipse. 

If a > b then PS + PS' = 2a = Major axis

If b > a then PS + PS' = 2b = Major axis

Calculation:

Given:

Equation of ellipse is 

Here a² = 4 and b² = 9

⇒ a = 2 and b = 3

b > a  so the major axis lies on y – axis with length 2b.

Now, sum of the focal distance = 2b = 2 × 3 = 6 units