Parabola, Ellipse and Hyperbola MCQ Quiz - Objective Question with Answer for Parabola, Ellipse and Hyperbola - Download Free PDF

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.

Calculation:

Given,

Hyperbola equation:

Divide both sides by 225 to obtain standard form:

Thus, and .

Compute from :

The foci are at , so the distance between them is :

∴ The distance between the two foci is  units.

Hence, the correct answer is Option 2.

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  one has 

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.

Concept:

Conic Sections and Rotated Parabolas:

• A conic section is the curve obtained by intersecting a cone with a plane. It includes parabolas, ellipses, and hyperbolas.
• A parabola is a conic section where the eccentricity e = 1.
• When the general second-degree equation Ax² + B x y + Cy² + Dx + E y + F = 0 has a non-zero B, it may represent a rotated conic.
• To eliminate the x y term, we perform a rotation of axes using:
  • x = x′cosθ − y′sinθ
  • y = x′sinθ + y′cosθ
• The angle of rotation θ is found using: tan(2θ) = B / (A − C)
• After rotation, the equation becomes a conic without the mixed term, allowing easy identification of axis, vertex, focus, etc.
• Vertex of the parabola is the point from which distances to the focus and directrix are equal.
• Latus rectum is the chord through the focus perpendicular to the axis of the parabola.

Calculation:

Given,

General equation: 3x² + 4xy + y² − 10x + 2y + 1 = 0

Let A = 3, B = 4, C = 1, D = -10, E = 2, F = 1

To eliminate xy term:

tan(2θ) = B / (A − C) = 4 / (3 − 1) = 2

⇒ 2θ = tan⁻¹(2)

⇒ θ = 1/2 × tan⁻¹(2)

Use θ to transform coordinates (x, y) → (x′, y′)

⇒ New equation becomes a standard parabola in rotated axes

⇒ Compare with standard form (y′ − k)² = 4a(x′ − h)

⇒ Vertex (h, k) = (1, 1)

⇒ Focus (x, y) = (7/2, −1/2) after back transformation

⇒ Directrix equation transforms to: x + y − 3 = 0

⇒ Axis of parabola: y = −x + 4

⇒ Equation of Latus rectum in transformed axes becomes x − 2y + 1 = 0

∴ The correct matches are:

A–R, B–Q, C–T, D–S

Concept:

Given:

The equation of a parabola is y2 + 12x = 0 

Tangent is drawn at point (3, 8)

Calculation:

y2 + 12x = 0 

⇒ y2 = - 12x

On comparing with the parabola equation, we get

⇒ 4a = - 12

⇒ a = - 3

On substituting on the equation of tangent in terms of slope, we get as

     --- equation (1)

Since the tangent passes through (3, 8) so, (3, 8) satisfy the equation (1)

⇒ 

⇒ 3m² - 8m - 3 = 0

⇒ m = 3, - 1/3

On substituting in the above equation (1), we get two tangents as,

⇒ x + 3y - 27 = 0 and 3x - y -1 = 0

So, only one option available so x + 3y - 27 = 0 is the required tangent.

Concept:

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)

• 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 =  = 

Concept:

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) =  

= 

= 

= 

CONCEPT:

The properties of a rectangular hyperbola  are:

• 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.

Concept

The equation of the hyperbola is  

The distance between the foci of a hyperbola = 2ae 

Again, 

Calculations:

The equation of the hyperbola is  ....(1)

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

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:

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).

Concept:

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.

Concept:

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)

Calculation:

Given:

16x² – 9y² = 1

Compare with 

∴ a² = 1/16 and b² = 1/9

Eccentricity =  

Concept:

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)

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)

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.