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

Explanation:

\(2 { b}=\frac{1}{4}(2{ae}) \)

\(\Rightarrow 4 {b}= {ae} \)

\(\Rightarrow 16 {b}^{2}={a}^{2} {e}^{2} \)  

\(\Rightarrow 16 {a}^{2}\left(1- {e}^{2}\right)={a}^{2} {e}^{2} \)  

\(\Rightarrow 16-16 {e}^{2}= {e}^{2} \)  

\(\Rightarrow {e}^{2}=\frac{16}{17} \)  

\(\Rightarrow {e}=\frac{4}{\sqrt{17}} \)  

hence Option 4 is the correct answer

Concept:

Ellipse Equation:

• The general equation of an ellipse is given by \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where a is the semi-major axis and b is the semi-minor axis.
• The equation of a circle is a special case of an ellipse where a = b.
• The distance between two points on the ellipse can be calculated using the distance formula.

Line Joining Two Points:

• The equation of a line joining two points \( P(x_1, y_1) \) and \( Q(x_2, y_2) \) is given by the slope-intercept form:
• \( y - y_1 = m(x - x_1) \), where m is the slope of the line.

Calculation:

Let \( P(x_1, y_1) \) and \( Q(x_2, y_2) \) be two distinct points on the ellipse:

\( P\left(\frac{3}{9}, 1\right) \) and \( Q\left(\frac{3}{2}, \sqrt{3}\right) \).

The equation of the line joining points P and Q is:

\( y - 1 = \frac{-\sqrt{3}}{2} \left(x - \frac{3}{2}\right) \)

Solving for the line's intersection with the x-axis:

\( N_2(x_2, 0) = \left(\frac{3}{2}, 0\right) \)

The equation of the line joining P and Q is:

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

Answer Options:

A) The equation of the line joining P and Q is \( y = 2x + 3y - 3(1 + \sqrt{3}) \), which is TRUE.

B) The equation of the line joining P and Q is \( y = 2x + 3y - 3(1 + \sqrt{3}) \), which is FALSE.

C) \( N_2(x_2, 0) = \left( \frac{3}{2}, 0 \right) \) is TRUE.

D) \( N_2(x_2, 0) = \left( 9 \times \frac{1}{2} \right) \) is FALSE.

Conclusion:

The correct answer is: Option A and Option C are TRUE, as per the above calculations.

Hence, the final value of \( |XY| = 3 \times \sqrt{3} \) is correct.

Calculation: 

Given that the ellipse is \(\frac{x^2}{36} + \frac{y^2}{25} = 1\)

Any point on line AB can be assumed as

⇒ \(Q\bigl(\sqrt{5}+r\cosθ,\;\sqrt{5}+r\sinθ\bigr).\)

Substitute into the ellipse equation

⇒ \(25\bigl(\sqrt{5}+r\cosθ\bigr)^2\;+\;36\bigl(\sqrt{5}+r\sinθ\bigr)^2 =900.\)

Simplifying

⇒ \(r^2\bigl(25\cos^2θ + 36\sin^2θ\bigr)\;+\;2\sqrt{5}\,r\,(25\cosθ + 36\sinθ)-595=0.\)

 Its roots are \(r = \pm PA,\pm PB\)

\(PA\cdot PB=\frac{|-595|}{25\cos^2θ + 36\sin^2θ}=\frac{595}{25\cos^2θ + 36\sin^2θ}. \)

⇒ \(PA\cdot PB = \frac{595}{25( 1- sin^2θ )+ 36 sin^2θ}= \frac{595}{25+11\sin^2θ}.\)

This is maximized when sin²θ  = 0

This means Line AB must be parallel to the x-axis

⇒ \(y_A=y_B=\sqrt5.\)

Putting \(y=\sqrt5\) in the equation of an ellipse 

\(\frac{x^2}{36} + \frac{(\sqrt5)^2}{25} = 1 \;\Longrightarrow\; \frac{x^2}{36} + \frac{5}{25}=1 \;\Longrightarrow\; \frac{x^2}{36}=\frac{4}{5}\)

\(\;\Longrightarrow\; x^2=\frac{36\cdot4}{5} \;\Longrightarrow\; x=\pm\frac{12}{\sqrt5}. \)

Hence,

\(A\!\bigl(-\tfrac{12}{\sqrt5},\sqrt5\bigr),\quad B\!\bigl(\tfrac{12}{\sqrt5},\sqrt5\bigr),\quad P\!\bigl(\sqrt5,\sqrt5\bigr).\)

⇒\(PA^2 =\Bigl(-\tfrac{12}{\sqrt5}-\sqrt5\Bigr)^2 +(\sqrt5-\sqrt5)^2 =\Bigl(\tfrac{-17}{\sqrt5}\Bigr)^2 =\frac{289}{5}, \)

⇒ \(PB^2 =\Bigl(\tfrac{12}{\sqrt5}-\sqrt5\Bigr)^2 +(\sqrt5-\sqrt5)^2 =\Bigl(\tfrac{7}{\sqrt5}\Bigr)^2 =\frac{49}{5}. \)

⇒ \(PA^2 + PB^2 = \frac{289}{5}+\frac{49}{5} = \frac{338}{5}\)

⇒ \(5\bigl(PA^2+PB^2\bigr) = 338.\)

Hence, the correct answer is option 4.

Calculation:

PS + PS' = 2 × 3 √2 

⇒ b² = a² (1 - e²) ⇒ 9 = 18( 1 - e²)

⇒ e = \(\frac{1}{\sqrt2}\)

Diretrix x = \(\frac{a}{e} = \frac{3\sqrt2}{\frac{1}{\sqrt2}{}} = 6\)

⇒ PS.PS' = \(PS.PS' = | \frac{1}{\sqrt{2}} (3\sqrt{2} \cos \theta - 6)|\)

 = \(\frac{1}{2}| 18 cos^2\theta - 36|\)

⇒ (PS. PS')_max = 18 and (PS .PS')_min = 9

Sum = 27

Hence, the correct answer is Option 4.

Calculation 

\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 foci are (ae, 0) and (–ae, 0) 

\(\frac{x^{2}}{A^{2}}+\frac{y^{2}}{B^{2}}\) = 1  foci are (Ae', 0) and (–Ae', 0) 

⇒ 2ae = \(2 \sqrt{3}\) ⇒ ae = \(\sqrt{3}\)

and 2Ae' = \(2 \sqrt{3}\) ⇒ Ae' = \(\sqrt{3}\)

⇒ ae = Ae' ⇒ \(\frac{\mathrm{e}}{\mathrm{e}^{\prime}}=\frac{\mathrm{A}}{\mathrm{a}}\)

⇒ \(\frac{1}{3}=\frac{\mathrm{A}}{\mathrm{a}}\) ⇒  a = 3A

Now a – A = 2 ⇒ a – \(\frac{\mathrm{a}}{3}\) – 2 ⇒ a = 3 and A = 1 

Ae =  \(\sqrt{3}\) ⇒ e = \(\frac{1}{\sqrt{3}}\) and e' = \(\sqrt{3}\)

b² = a²(1 – e²)

b² = 6

and B² = A²((e')² – 1) = (2) ⇒ B² = 2

sum of LR = \(\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}+\frac{2 \mathrm{~B}^{2}}{\mathrm{~A}}\) = 8

Hence option 3 is correct

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: 

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. 

Concept:

Tangent to an Ellipse:

The equation of the tangent to the ellipse \(\rm \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\), at a point (x₁, y₁), is given by: \(\rm \frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1\).

Calculation:

At x = 3, we will have:

\(\rm \frac{3^2}{25}+\frac{y^2}{16}=1\)

⇒ \(\rm \frac{y^2}{16}={16\over25}\)

⇒ y = ± \(16\over5\)

From the above formula, we can say that the required equation of the tangent to the ellipse \(\rm \frac{x^2}{25}+\frac{y^2}{16}=1\) at \(\left(3,{16\over5}\right)\) and \(\left(3,{-16\over5}\right)\) are (respectively):

\(\rm \frac{x(3)}{25}+\frac{y\left({16\over5}\right)}{16}=1\)

⇒ \(\rm \frac{3x}{25}+\frac{y}{5}=1\)

⇒ 3x + 5y = 25

OR

\(\rm \frac{x(3)}{25}+\frac{y\left({-16\over5}\right)}{16}=1\)

⇒ \(\rm \frac{3x}{25}-\frac{y}{5}=1\)

⇒ 3x - 5y = 25

Concept:

The standard equation of an ellipse:

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

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

The eccentricity = \(\rm \sqrt{(a^2-b^2)}\over a\)

Length of latus rectum = \(\rm 2b^2 \over a\)

Distance from center to focus = \(\rm \sqrt{a^2-b^2}\)

Calculation:

Given ellipse \(\rm {x^2\over100}+{y^2\over64} = 1\)

a² = 100 ⇒ a = 10

and b² = 64 ⇒ b = 8

The eccentricity (e)

⇒ e = \(\rm \sqrt{100-64}\over 10\)

⇒ e = \(\rm \sqrt{36}\over 10\)

⇒ e = \(\rm 6\over 10\)

⇒ e = 0.6

Now distance between foci = 2ae

= 2 × 10 × 0.6

∴ Distance between foci = 12

Concept:

Calculation:

25x² + 16y² = 400

\( \Rightarrow \frac{{25{{\rm{x}}^2}}}{{400}} + \frac{{16{{\rm{y}}^2}}}{{400}} = 1\)

\( \Rightarrow \frac{{{{\rm{x}}^2}}}{{16}} + \frac{{{{\rm{y}}^2}}}{{25}} = 1\)

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

Since ( a < b )

Concept:

The length of the latus rectum of the ellipse \(\rm \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) is equal to \(\rm \frac{2a^2}{b}\). (a < b)

Calculation:

Writing the equation of the ellipse in the standard form \(\rm \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\), we get:

\(\rm \frac{x^2}{2^2}+\frac{y^2}{(2\sqrt3)^2}=1\)

∴ a = 2 and b = 2√3.

Here a < b

Length of the latus rectum = \(\rm \frac{2a^2}{b}\) = \(\rm \frac{2(2^2)}{2\sqrt3}\) = \(\rm \frac{4}{\sqrt3}\).

Concept:

The general equation of the ellipse is:

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

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)

\(⇒ b^{2}=81\left ( 1-\frac{1}{9} \right )\)

⇒ b² = 72

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

\(\frac{x^2}{81}+\frac{y^2}{72}=1\)

Hence, the equation of the ellipse is \(\frac{x^2}{81}+\frac{y^2}{72}=1\).

Concept:

1. Equation of circle x2 + y2 = r2

2. Equation of an ellipse \(\frac{{{x^2}}}{{{a^2}}}\; + \frac{{{y^2}}}{{{b^2}}} = 1\)

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

4. Equation of hyperbola \(\frac{{{x^2}}}{{{a^2}}} - \;\frac{{{y^2}}}{{{b^2}}} = 1\)

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) 

\({\left( {\frac{x}{3}} \right)^2} = 1 + sin2t\) ----(1)

\({\left( {\frac{y}{4}} \right)^2} = 1 - sin2t\) ----(2)

Adding r=equation 1 and 2;

\(\frac{{{x^2}}}{9} + \frac{{{y^2}}}{{16}} = 2\)  

Hence, the given curve represents an ellipse.

Concept:

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

• 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 an ellipse: \(\frac{{{\rm{\;}}{{\bf{x}}^2}}}{{{{\bf{a}}^2}}} + \frac{{{{\bf{y}}^2}}}{{{{\bf{b}}^2}}} = 1\)

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:

The standard equation of an ellipse is given by: \(\frac{{{\rm{\;}}{{\bf{x}}^2}}}{{{{\bf{a}}^2}}} + \frac{{{{\bf{y}}^2}}}{{{{\bf{b}}^2}}} = 1\) 

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 \(\rm \dfrac{x^2}{4}+\dfrac{y^2}{9}=1\)

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