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,

\(x = a\,\sinα,\quad y = b\,\cosα\)

we identify

\(a = 3,\quad b = 5 \)

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

\(e = \sqrt{\,1 - \frac{(\text{semi-minor})^{2}}{(\text{semi-major})^{2}}\,} = \sqrt{\,1 - \frac{a^{2}}{b^{2}}\,} \)

Substitute (a = 3) and (b = 5):

\(e = \sqrt{\,1 - \frac{3^{2}}{5^{2}}\,} = \sqrt{\,1 - \frac{9}{25}\,} = \sqrt{\frac{16}{25}} = \frac{4}{5} \)

Hence, the correct answer is Option 2.

Calculation:

Given,

Hyperbola equation: \(25x^{2} - 75y^{2} = 225\)

Divide both sides by 225 to obtain standard form:

\(\frac{25x^{2}}{225} - \frac{75y^{2}}{225} = 1 \;\Longrightarrow\; \frac{x^{2}}{9} - \frac{y^{2}}{3} = 1\)

Thus, \(a^{2} = 9\) and \(b^{2} = 3\).

Compute \(c\) from \(c^{2} = a^{2} + b^{2}\):

\(c^{2} = 9 + 3 = 12 \;\Longrightarrow\; c = \sqrt{12} = 2\sqrt{3}.\)

The foci are at \((\pm c,\,0)\), so the distance between them is \(2c\):

\(2c = 2 \times 2\sqrt{3} = 4\sqrt{3}.\)

∴ The distance between the two foci is \(4\sqrt{3}\) 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 

\(m \;=\; \tan(45^\circ) \;=\; 1.\)

A standard parametric form for y2 = 4x is 

\(\bigl(x(t),\,y(t)\bigr) \;=\;\bigl(t^{2},\,2t\bigr), \)

since  \(y^{2} = 4x \implies (2t)^{2} = 4\,t^{2} \)

The slope of the tangent at \(\bigl(t^{2},\,2t\bigr) \)

Differentiate y2 = 4x 

\(2y\,\frac{dy}{dx} \;=\; 4 \)

\(\;\Longrightarrow\; \frac{dy}{dx} \;=\; \frac{4}{\,2y\,} \;=\; \frac{2}{\,y\,}. \)

At the point \(\bigl(x,y\bigr) = \bigl(t^{2},\,2t\bigr) \) one has \(y = 2t \)

\(\left.\frac{dy}{dx}\right|_{\,(t^{2},\,2t)} \) \(= \frac{2}{\,2t\,} = \frac{1}{t}.\)

We require this slope to equal 1.

\(\frac{1}{t} \;=\; 1 \;\Longrightarrow\; t = 1. \)

Now point of contact

Substitute t = 1 in \(\bigl(x(t),\,y(t)\bigr) = \bigl(t^{2},\,2t\bigr)\). 

\(x(1) = 1^{2} = 1, \)

\(y(1) = 2 \cdot 1 = 2. \)

Thus the point of contact of the tangent of slope 1 is (1, 2)

Hence, the correct answer is Option 4.

Calculation: 

Concept:

• Hyperbola Standard Equation: If the vertices are on the x-axis at (±a, 0), the equation is (x²/a²) - (y²/b²) = 1.
• Vertices: Given as (±6, 0), so a = 6 ⇒ a² = 36.
• Eccentricity: e = √5 / 2. For a hyperbola, e = √(1 + b²/a²).
• Normal: A line perpendicular to the tangent at a point on the curve. The slope of the normal is the negative reciprocal of the tangent slope.
• The given normal is parallel to the line √2x + y = 2√2, hence the slope is -√2.

Calculation:

Given:

a = 6 ⇒ a² = 36,

and e = √5 / 2

⇒ e² = 5 / 4

⇒ 5 / 4 = 1 + b² / 36 ⇒ b² = 9

So the hyperbola is: (x² / 36) - (y² / 9) = 1

Differentiating implicitly:

(2x / 36) - (2y / 9) × (dy/dx) = 0 ⇒ x / 18 = (2y / 9)(dy/dx)

⇒ dy/dx = x / 4y

⇒ slope of tangent = x / 4y

⇒ slope of normal = -4y / x

Given: slope of normal = -√2 ⇒ -4y / x = -√2 ⇒ 4y / x = √2

⇒ y = (x√2) / 4

Substitute into hyperbola:

(x² / 36) - (1 / 9) × ((x√2 / 4)²) = 1

⇒ (x² / 36) - (1 / 9) × (2x² / 16) = 1

⇒ (x² / 36) - (x² / 72) = 1

⇒ (x²)(1/36 - 1/72) = 1 ⇒ x² × (1/72) = 1 ⇒ x² = 72

⇒ x = 6√2

Then, y = (x√2) / 4 = (6√2 × √2) / 4 = 12 / 4 = 3

So, the point on the hyperbola is (6√2, 3)

The equation of the normal line with slope -√2 passing through (6√2, 3):

y - 3 = -√2(x - 6√2)

To find the y-intercept, put x = 0:

y = 3 + √2 × 6√2 = 3 + 12 = 15

So, line segment lies between (6√2, 3) and (0, 15)

Length d = √[(6√2)² + (15 - 3)²] = √[72 + 144] = √216

∴ The required value of d² is 216.

Concept:

Focal Chord of Parabola and Circle Properties:

• The focal chord of the parabola \( y^2 = 4ax \)passes through points on the parabola making a certain angle with the x-axis.
• The circle with one diameter as the focal chord touches the y-axis at a certain point.

• The tangent of the given angle and the coordinates of the point P on the parabola can be used to find the parameter t.
• The equation of the circle with diameter PS and its intersection with the y-axis is used to find \(\alpha\)

Calculation:

Given,

The parabola: \( y^2 = 4x \)

Focal chord PQ makes an angle of \(60^\circ\) with positive x-axis.

Using slope\( m = \tan 60^\circ = \sqrt{3}\)

\( \tan 60^\circ = \frac{2t - 0}{t^2 - 1} = \sqrt{3} \implies t = \sqrt{3} \)

Coordinates of P: \( (3, 2\sqrt{3}) \)

Circle with diameter PS and focus S:

\( (x - 1)(x - 3) + (y - 0)(y - 2\sqrt{3}) = 0 \)

At x = 0,

\( (-1)(-3) + y(y - 2\sqrt{3}) = 0 \implies 3 + y^2 - 2\sqrt{3}y = 0 \implies (y - \sqrt{3})^2 = 0 \implies y = \sqrt{3} = \alpha \)

Calculate \( 5\alpha^2:\)

\( 5 \alpha^2 = 5 (\sqrt{3})^2 = 5 \times 3 = 15 \)

∴ The correct answer is Option 1.

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.