Parabola, Ellipse and Hyperbola MCQ Quiz - Objective Question with Answer for Parabola, Ellipse and Hyperbola - Download Free PDF
Last updated on Jul 4, 2025
Latest Parabola, Ellipse and Hyperbola MCQ Objective Questions
Parabola, Ellipse and Hyperbola Question 1:
If any point on an ellipse is (3sin
Answer (Detailed Solution Below)
Parabola, Ellipse and Hyperbola Question 1 Detailed Solution
Calculation:
Given any point on the ellipse is 3sinα, 5cosα. In the standard parametric form of an ellipse,
we identify
Since (b > a), the semi-major axis is (b = 5) and the semi-minor axis is (a = 3). The eccentricity e of an ellipse is
Substitute (a = 3) and (b = 5):
Hence, the correct answer is Option 2.
Parabola, Ellipse and Hyperbola Question 2:
. What is the distance between the two foci of the hyperbola 25x2 - 75y 2= 225 ?
Answer (Detailed Solution Below)
Parabola, Ellipse and Hyperbola Question 2 Detailed Solution
Calculation:
Given,
Hyperbola equation:
Divide both sides by 225 to obtain standard form:
Thus,
Compute
The foci are at
∴ The distance between the two foci is
Hence, the correct answer is Option 2.
Parabola, Ellipse and Hyperbola Question 3:
A tangent to the parabola y2 = 4x is inclined at an angle 45° deg with the positive direction of x-axis. What is the point of contact of the tangent and the parabola?
Answer (Detailed Solution Below)
Parabola, Ellipse and Hyperbola Question 3 Detailed Solution
Calculation:
Given the parabola
y2 = 4x
and a tangent to this parabola that is inclined at an angle of 45°with the positive x-axis, Hence, its slope is
A standard parametric form for y2 = 4x is
since
The slope of the tangent at
Differentiate y2 = 4x
At the point
We require this slope to equal 1.
Now point of contact
Substitute t = 1 in
Thus the point of contact of the tangent of slope 1 is (1, 2)
Hence, the correct answer is Option 4.
Parabola, Ellipse and Hyperbola Question 4:
Let T and C respectively be the transverse and conjugate axes of the hyperbola 16x2 - y2 + 64x + 4y + 44 = 0. Then the area of the region above the parabola x2 = y + 4, below the transverse axis T and on the right of the conjugate axis C is:
Answer (Detailed Solution Below)
Parabola, Ellipse and Hyperbola Question 4 Detailed Solution
Calculation:
16(x2 + 4x) – (y2 – 4y) + 44 = 0
⇒ 16(x + 2)2 – 64 – (y – 2)2 + 4 + 44 = 0
⇒ 16(x + 2)2 – (y – 2)2 = 16
⇒
⇒
⇒
⇒
⇒
Hence, the correct answer is Option 2.
Parabola, Ellipse and Hyperbola Question 5:
The equations of two sides of a variable triangle are x = 0 and y = 3, and its third side is a tangent to the parabola y2 = 6x. The locus of its circumcentre is :
Answer (Detailed Solution Below)
Parabola, Ellipse and Hyperbola Question 5 Detailed Solution
Calculation:
y2 = 6x & y2 = 4ax
⇒
⇒
⇒ 3h = 2(–2k2 + 9k – 9)
⇒ 4y2 – 18y + 3x + 18 = 0
Hence, the correct answer is Option 3.
Top Parabola, Ellipse and Hyperbola MCQ Objective Questions
The length of latus rectum of the hyperbola
Answer (Detailed Solution Below)
Parabola, Ellipse and Hyperbola Question 6 Detailed Solution
Download Solution PDFConcept:
Hyperbola: The locus of a point which moves such that its distance from a fixed point is greater than its distance from a fixed straight line. (Eccentricity = e > 1)
Equation |
|
|
Equation of Transverse axis |
y = 0 |
x = 0 |
Equation of Conjugate axis |
x = 0 |
y = 0 |
Length of Transverse axis |
2a |
2b |
Length of Conjugate axis |
2b |
2a |
Vertices |
(± a, 0) |
(0, ± b) |
Focus |
(± ae, 0) |
(0, ± be) |
Directrix |
x = ± a/e |
y = ± b/e |
Centre |
(0, 0) |
(0, 0) |
Eccentricity |
|
|
Length of Latus rectum |
|
|
Focal distance of the point (x, y) |
ex ± a |
ey ± a |
- Length of Latus rectum =
Calculation:
Given:
Compare with the standard equation of a hyperbola:
So, a2 = 100 and b2 = 75
∴ a = 10
Length of latus rectum =
The eccentricity of the hyperbola
Answer (Detailed Solution Below)
Parabola, Ellipse and Hyperbola Question 7 Detailed Solution
Download Solution PDFConcept:
Standard equation of an hyperbola :
- Coordinates of foci = (± ae, 0)
- Eccentricity (e) =
⇔ a2e2 = a2 + b2 - Length of Latus rectum =
Calculation:
Given:
Compare with the standard equation of a hyperbola:
So, a2 = 100 and b2 = 75
Now, Eccentricity (e) =
=
=
=
Find the equation of the hyperbola, the length of whose latus rectum is 4 and the eccentricity is 3 ?
Answer (Detailed Solution Below)
Parabola, Ellipse and Hyperbola Question 8 Detailed Solution
Download Solution PDFCONCEPT:
The properties of a rectangular hyperbola
- Its centre is given by: (0, 0)
- Its foci are given by: (- ae, 0) and (ae, 0)
- Its vertices are given by: (- a, 0) and (a, 0)
- Its eccentricity is given by:
- Length of transverse axis = 2a and its equation is y = 0.
- Length of conjugate axis = 2b and its equation is x = 0.
- Length of its latus rectum is given by:
CALCULATION:
Here, we have to find the equation of hyperbola whose length of latus rectum is 4 and the eccentricity is 3.
As we know that, length of latus rectum of a hyperbola is given by
⇒
⇒ b2 = 2a
As we know that, the eccentricity of a hyperbola is given by
⇒ a2e2 = a2 + b2
⇒ 9a2 = a2 + 2a
⇒ a = 1/4
∵ b2 = 2a
⇒ b2 = 1/2
So, the equation of the required hyperbola is 16x2 - 2y2 = 1
Hence, option B is the correct answer.
The distance between the foci of a hyperbola is 16 and its eccentricity is √2. Its equation is
Answer (Detailed Solution Below)
Parabola, Ellipse and Hyperbola Question 9 Detailed Solution
Download Solution PDFConcept
The equation of the hyperbola is
The distance between the foci of a hyperbola = 2ae
Again,
Calculations:
The equation of the hyperbola is
The distance between the foci of a hyperbola is 16 and its eccentricity e = √2.
We know that The distance between the foci of a hyperbola = 2ae
⇒ 2ae = 16
⇒ a =
Again,
⇒
⇒
Equation (1) becomes
⇒
⇒ x2 - y2 = 32
The equation of the ellipse whose vertices are at (± 5, 0) and foci at (± 4, 0) is
Answer (Detailed Solution Below)
Parabola, Ellipse and Hyperbola Question 10 Detailed Solution
Download Solution PDFConcept:
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 ⇒
ae = 4 ⇒ e = 4/5
Now, 4/5 =
∴ Equation of ellipse =
Hence, option (1) is correct.
The vertex of the parabola (y - 3)2 = 20(x - 1) is:
Answer (Detailed Solution Below)
Parabola, Ellipse and Hyperbola Question 11 Detailed Solution
Download Solution PDFConcept:
Standard Form of the equation: | (y - k)2 = 4a(x - h) |
Equation of the Axis: | y = k |
Vertex: | (h, k) |
Focus: | (h + a, k) |
Directrix: | x = h - a |
Calculation:
Comparing the given equation (y - 3)2 = 20(x - 1) with the general equation of the parabola (y - k)2 = 4a(x - h), we can say that:
k = 3, a = 5, h = 1.
Vertex is (h, k) = (1, 3).
In the parabola y2 = x, what is the length of the chord passing through the vertex and inclined to the x-axis at an angle θ?
Answer (Detailed Solution Below)
Parabola, Ellipse and Hyperbola Question 12 Detailed Solution
Download Solution PDFConcept:
The coordinates of the point where the chord cut the parabola satisfIes the equation of a parabola.
Calculation:
Given:
The equation of a parabola is y2 = x.
The angle made by Chord OA with x-axis is θ
Let the length of the chord OA of the parabola is L
So, Length of AM = L sinθ
and Length of OM = L cosθ
So, The coordinate of A = (L cos θ, L sin θ)
And this point will satisfy the equation of parabola y2 = x.
⇒ (Lsin θ)2 = L cos θ
⇒L2 sin2 θ = L cos θ
⇒ L = cos θ. cosec2 θ
∴ The required length of chord is cos θ. cosec2 θ.
The eccentricity of the hyperbola 16x2 – 9y2 = 1 is
Answer (Detailed Solution Below)
Parabola, Ellipse and Hyperbola Question 13 Detailed Solution
Download Solution PDFConcept:
Hyperbola: The locus of a point which moves such that its distance from a fixed point is greater than its distance from a fixed straight line. (Eccentricity = e > 1)
Equation |
|
|
Equation of Transverse axis |
y = 0 |
x = 0 |
Equation of Conjugate axis |
x = 0 |
y = 0 |
Length of Transverse axis |
2a |
2b |
Length of Conjugate axis |
2b |
2a |
Vertices |
(± a, 0) |
(0, ± b) |
Focus |
(± ae, 0) |
(0, ± be) |
Directrix |
x = ± a/e |
y = ± b/e |
Centre |
(0, 0) |
(0, 0) |
Eccentricity |
|
|
Length of Latus rectum |
|
|
Focal distance of the point (x, y) |
ex ± a |
ey ± a |
Calculation:
Given:
16x2 – 9y2 = 1
Compare with
∴ a2 = 1/16 and b2 = 1/9
Eccentricity =
What is the focus of the parabola x2 = 16y ?
Answer (Detailed Solution Below)
Parabola, Ellipse and Hyperbola Question 14 Detailed Solution
Download Solution PDFConcept:
Parabola: The locus of a point which moves such that its distance from a fixed point is equal to its distance from a fixed straight line. (Eccentricity = e =1)
Equation |
x2 = 4ay; |
Vertex |
(0, 0) |
Focus |
(0, a) |
Equation of the directrix |
y = -a |
Equation of the axis |
x = 0 |
Length of Latus rectum |
4a |
Focal distance |
y + a |
Calculation:
Given: x2 = 16y
⇒ x2 = 4 × 4 × y
Compare with standard equation of parabola x2 = 4ay
So, a = 4
Therefore, Focus = (0, a) = (0, 4)
Length of Latus rectum of ellipse
Answer (Detailed Solution Below)
Parabola, Ellipse and Hyperbola Question 15 Detailed Solution
Download Solution PDFConcept:
Standard equation of ellipse ,
Length of latus rectum , L.R =
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.