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

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

\(y = mx + \frac{-3}{m}\)     --- equation (1)

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

⇒ \(8 = 3m + \frac{-3}{m}\)

⇒ 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:

The eccentricity of the hyperbola \({x^2 \over a^2} -{y^2 \over b^2}=1 \) is \(\sqrt{1+{b^2 \over a^2 }} \)

The equation of the tangent to the hyperbola \({x^2 \over a^2} -{y^2 \over b^2}=1 \) at (h,k) is

\({hx \over a^2} -{ky \over b^2}=1 \)

Calculation:

Let the equation of hyperbola be \({x^2 \over a^2} -{y^2 \over b^2}=1 \)

⇒ Eccentricity = 2 = \(\sqrt{1+{b^2 \over a^2 }}\)

⇒ 3a² = b²

And hyperbola passes through (4,6)

⇒ \({4^2 \over a^2} -{6^2 \over b^2}=1 \)

⇒ \({16 \over a^2} -{36 \over 3a^2}=1 \)

⇒ a² = 4 and b² = 12

⇒ The equation of the tangent to the hyperbola \({x^2 \over 4} -{y^2 \over 12}=1 \) at (4,6) is

\({4x \over 4} -{6y \over 12}=1 \)

⇒ 2x - y - 2 = 0

∴ The correct answer is option (4).

Concept:

Equation of ellipse: \(\rm\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)

Eccentricity = \(\rm\sqrt{1-\frac{b^2}{a^2}}\)

Calculation:

Here, 4x2 + 9y2 = 144

\(\Rightarrow \rm \frac{4x^2}{144}+\frac{9y^2}{144}=1\\ \Rightarrow \rm \frac{x^2}{\frac{144}{4}}+\frac{y^2}{\frac{144}{9}}=1\\ \)

So, a² = 144/4 and b² = 144/9

∴Eccentricity

 \(\rm\sqrt{1-\frac{b^2}{a^2}} \\=\sqrt{1-\frac{144}{9}\times \frac{4}{144}}\\ =\sqrt{1-\frac{4}{9}}\\ =\sqrt{\frac{5}{9}}\\ =\frac{\sqrt5}{3}\)

Hence, option (1) is correct. 

Concept:

1. Position of a point with respect to a curve:

• For any curve F(x, y) = k in the xy-plane with k is a real number, then the point (a, b) lies on the curve if F(a, b) = k.
• For any curve F(x, y) = k in the xy-plane with k is a real number, then the point (a, b) lies outside the curve if F(a, b) > k.
• For any curve F(x, y) = k in the xy-plane with k is a real number, then the point (a, b) lies inside the curve if F(a, b) < k.

2. Standard equation of an ellipse and its foci:

• The standard equation of an ellipse is of the form \(\frac{{{{\rm{x}}^2}}}{{{{\rm{a}}^2}}} + \frac{{{{\rm{y}}^2}}}{{{{\rm{b}}^2}}} = 1\). If a² > b² then the ellipse is horizontal and vertical otherwise.
• The foci of the ellipse is given by (±c, 0) if the ellipse is horizontal and it is given by (0, ±c) if the ellipse is vertical.

Solution:

Observe that the given equation is of the form F(x, y) = k, where F(x, y) = 2x² + 7y² and k = 20.

Therefore, to find the position of the given point (1, 2) with respect to ellipse substitute x = 1 and y = 2 in the equation of the ellipse..

Therefore,

\(2{\left( 1 \right)^2} + {\rm{\;}}7{\left( 2 \right)^2} = 30\)

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 = \(\rm \frac{2b^2}{a}\)

Calculation:

Given: \(\rm \frac{x^2}{100} - \frac{y^2}{75} = 1\)

Compare with the standard equation of a hyperbola: \(\frac{{{\rm{\;}}{{\bf{x}}^2}}}{{{{\bf{a}}^2}}} - \frac{{{{\bf{y}}^2}}}{{{{\bf{b}}^2}}} = 1\)

So, a2 = 100 and b2 = 75

∴ a = 10

Length of latus rectum =  \(\rm \frac{2b^2}{a}\)= \(\rm \frac{2 \times 75}{10} = 15\)

Concept:

Standard equation of an hyperbola : \(\frac{{{\rm{\;}}{{\bf{x}}^2}}}{{{{\bf{a}}^2}}} - \frac{{{{\bf{y}}^2}}}{{{{\bf{b}}^2}}} = 1\) 

• Coordinates of foci = (± ae, 0)
• Eccentricity (e) = \(\sqrt {1 + {\rm{\;}}\frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2}}}} \) ⇔ a2e2 = a2 + b2
• Length of Latus rectum = \(\rm \frac{2b^2}{a}\)

Calculation:

Given: \(\rm \frac{x^2}{100} - \frac{y^2}{75} = 1\)

Compare with the standard equation of a hyperbola: \(\frac{{{\rm{\;}}{{\bf{x}}^2}}}{{{{\bf{a}}^2}}} - \frac{{{{\bf{y}}^2}}}{{{{\bf{b}}^2}}} = 1\)

So, a2 = 100 and b2 = 75

Now, Eccentricity (e) = \(\sqrt {1 + {\rm{\;}}\frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2}}}} \) 

= \(\sqrt {1 + \frac{75}{100}}\)

= \(\sqrt {1 + \frac{3}{4}}\)

= \(\sqrt { \frac{7}{4}}\)

CONCEPT:

The properties of a rectangular hyperbola \(\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\) 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: \(e = \frac{{\sqrt {{a^2} + {b^2}} }}{a}\)
• 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: \(\frac{2b^2}{a}\)

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 \(\frac{2b^2}{a}\)

⇒ \(\frac{2b^2}{a} = 4\)

⇒ b² = 2a

As we know that, the eccentricity of a hyperbola is given by \(e = \frac{{\sqrt {{a^2} + {b^2}} }}{a}\)

⇒ 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 \(\rm \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\) 

The distance between the foci of a hyperbola = 2ae 

Again, \(\rm b^2 = a^2(e^2-1)\)

Calculations:

The equation of the hyperbola is \(\rm \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\) ....(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  =  \(\dfrac {16}{2\sqrt 2}\) = \({4\sqrt 2}\)

Again, \(\rm b^2 = a^2(e^2-1)\)

⇒ \(\rm b^2 = 32(2-1)\)

⇒ \(\rm b^2 = 32\)

Equation (1) becomes

⇒ \(\rm \dfrac{x^2}{32} - \dfrac{y^2}{32} = 1 \)

⇒ x2 - y2 = 32

Concept:

Equation of ellipse: \(\rm\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)​

Eccentricity (e) = \(\rm\sqrt{1-\frac{b^2 }{a^2}}\)

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

Calculation:

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

So, a = ±5 ⇒ \(a^2=25\) and

ae = 4 ⇒ e = 4/5

Now, 4/5 = \(\rm\sqrt{1-\frac{b^2 }{5^2}}\)

\(⇒ \rm\frac{16}{25}=\rm\frac{25-b^2}{25}\\⇒ 16=25-b^2 \\⇒ b^2=9 \)

∴ Equation of ellipse = \(\rm \frac {x^2}{25} + \frac {y^2}{9} = 1\)

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

\( \Rightarrow \frac{{{\rm{\;}}{{\rm{x}}^2}}}{{\frac{1}{{16}}}} - \frac{{{{\rm{y}}^2}}}{{\frac{1}{9}}} = 1\)

Compare with \(\frac{{{\rm{\;}}{{\rm{x}}^2}}}{{{{\rm{a}}^2}}} - \frac{{{{\rm{y}}^2}}}{{{{\rm{b}}^2}}} = 1\)

∴ a² = 1/16 and b² = 1/9

Eccentricity = \(\sqrt {1 + {\rm{\;}}\frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2}}}} = \;\sqrt {1 + \;\frac{{\left( {\frac{1}{9}} \right)}}{{\left( {\frac{1}{{16}}} \right)}}} = \;\sqrt {1 + \;\frac{{16}}{9}} = \;\sqrt {\frac{{25}}{9}} = \;\frac{5}{3}\) 

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 , \(\rm\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}= 1\) 

Length of latus rectum , L.R = \(\rm\frac{2a^{2}}{b}\) , if  b > a

Calculation: 

\(\rm\frac{x^{2}}{25}+\frac{y^{2}}{49}= 1\) ,

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

We know that , Length of latus rectum = \(\rm\frac{2a^{2}}{b}\)  

⇒ L.R = \(\rm\frac{2\times5^{2}}{7}\) =  \(\rm\frac{50}{7}\) . 

The correct option is 2.