Latus Rectum MCQ Quiz in తెలుగు - Objective Question with Answer for Latus Rectum - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Calculation :

⇒ Given hyperbola: \( H_n \Rightarrow \frac{x^2}{1 + n} - \frac{y^2}{3 + n} = 1 \)

⇒ Eccentricity: \( e = \sqrt{1 + \frac{b^2}{a^2}} \)

⇒ \( e = \sqrt{1 + \frac{3 + n}{1 + n}} = \sqrt{\frac{2n + 4}{n + 1}} \)

⇒ For \( e \in \mathbb{Q} \), the expression under square root must be a perfect square.

⇒ Try smallest even \( n \) such that the value is rational: Let \( n = 48 \Rightarrow e = \sqrt{\frac{2(48) + 4}{48 + 1}} = \sqrt{\frac{100}{49}} = \frac{10}{7} \)

⇒ So, \( a^2 = 49, \quad b^2 = 51, \quad a = 7 \)

⇒ Length of Latus Rectum:

⇒ \( l = \frac{2b^2}{a} = \frac{2 \cdot 51}{7} = \frac{102}{7} \)

⇒ Multiply by 21: \( 21l = 21 \cdot \frac{102}{7} = 3 \cdot 102 = 306 \)

Hence, the correct answer is 306.

Calculation: 

\( \begin{align*} e_H &= \sqrt{2} \end{align*} \)

\( \begin{align*} e_E &= \frac{1}{\sqrt{2}} \end{align*} \)

Since the curves intersect each other orthogonally

The ellipse and the hyperbola are confocal

⇒ \( \begin{align*} H: \frac{x^2}{1/2} - \frac{y^2}{1/2} &= 1 \end{align*} \)

\( \begin{align*} \Rightarrow \text{foci} &= (1,0) \end{align*} \)

For ellipse ae_E = 1

\( \begin{align*} \Rightarrow a &= \sqrt{2} \end{align*} \)

⇒\( \begin{align*} (e_E)^2 &= \frac{1}{2} \Rightarrow 1 - \frac{b^2}{a^2} = \frac{1}{2} \Rightarrow \frac{b^2}{a^2} = \frac{1}{2} \end{align*} \)

\( \begin{align*} \Rightarrow b^2 &= 1 \end{align*} \)

 Length of L.R \(= \frac{2b^2}{a} = \frac{2}{\sqrt{2}} = \sqrt{2} \)

Hence, the correct answer is 2.

Answer (1) 

Sol. 

2b = \(\frac{1}{4}\)(ae) ⇒ 4b = ae

b² = a² - a²e²

b2 = a2 - 16b2

17b2 = a2

e = \(\sqrt{\frac{1-b^{2}}{a^{2}}}=\sqrt{\frac{1-1}{17}}=\frac{4}{\sqrt{17}}\)

Calculation: 

Length of LR = \(\rm \frac{2 b^{2}}{a}=10 \Rightarrow 5 a=b^{2} \) .....(1)

\(\rm f(t)=t^{2}+t+\frac{11}{12} \)

⇒ \(\rm \frac{d f(t)}{d t}=2 t+1=0 \Rightarrow t=\frac{-1}{2} \)

Min value of \(\rm f(t)=\left(\frac{-1}{2}\right)^{2}+\left(\frac{-1}{2}\right)+\frac{11}{12} \) 

\(=\frac{1}{4} \frac{-1}{2}+\frac{11}{12}=\frac{3-6+11}{12}=\frac{8}{12}=\frac{2}{3}=e \)

⇒ \(e^{2}=\frac{1-b^{2}}{a^{2}} \Rightarrow \frac{4}{9}=\frac{1-b^{2}}{a^{2}} \)

\(\rm \Rightarrow \frac{b^{2}}{a^{2}}=\frac{1-4}{a}=\frac{5}{a} \Rightarrow b^{2}=\frac{5 a^{2}}{a} \) ......(2)

From (1) & (2)

⇒ \(\rm 5 a=\frac{5 a^{2}}{a} \Rightarrow a=9, b=\sqrt{45}=3 \sqrt{5}\)

∴ a² + b² = 81 +45 = 126

Hence, the correct answer is Option 2.

Calculation:

 

Center C = \((\frac{2+2}{2}, \frac{5 +(-3)}{2}) = (2,1)\)

Distance between foci: 

⇒ 2c = \(\sqrt{(2-2)^2 + (5-(-3))^2} = 8 \)

⇒ c = 4

Also e = a/c ⇒ a = c/e = \(\frac{4}{\frac{4}{5}} = 5\)

Now b² = a² − c² = 25 − 16 = 9, 

⇒ b = 3

Length of latus rectum =  \(\frac{2b^2}{a} = \frac{2 \times 9}{5} = \frac{18}{5}\)

Hence, the correct answer is Option 4.

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 = \(\rm \frac {2b^2}{a}\)

Length of major axis =2a and Length of minor axis = 2b

Calculation: 

Here,  e = 4/5 =\(\sqrt {1 - {\rm{\;}}\frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2}}}} \)

Squaring both sides, we get

16/25 = \(\rm 1 - \frac {b^2}{a^2}\)

∴ (b2 / a2) = 1 - 16/25 =  9/25 ....(1)

Latus rectus 2b2 / a = 14.4

⇒ b2 / a = 7.2

Puting above value in (1),

 7.2/ a = 9/25

⇒ a = 20

Now, b2 = 7.2 × 20 = 144

⇒ b = 12

Sum of major and minor axes = 2a + 2b

= 2(20) + 2(12)

= 64 

Hence, option (3) is correct.

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:

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

Length of latus rectum = 2b²/a, when a > b and 2a²/b, when a < b

Calculation:

3x² + y² -12x + 2y + 1 = 0

⇒ 3(x² - 4x + 4) – 12 + (y² + 2y + 1) = 0

⇒ 3(x – 2)² – 12 + (y + 1)² = 0

⇒ 3(x – 2)² + (y + 1)² = 12

\( \Rightarrow \frac{{3{{\left( {{\rm{x}} - 2} \right)}^2}}}{{12}} + \frac{{{{\left( {{\rm{y}} + 1} \right)}^2}}}{{12}} = 1\)                                  (Divide by 12)

\( \Rightarrow \frac{{{{\left( {{\rm{x}} - 2} \right)}^2}}}{4} + \frac{{{{\left( {{\rm{y}} + 1} \right)}^2}}}{{12}} = 1\)

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

∴ a² = 2² and b² = (2√3)²

Here a < b

So, length of latus rectum = 2a²/b

= \(\frac{{2\left( 4 \right)}}{{2\sqrt 3 }}\)

= \(\frac{4}{\sqrt 3}\) units

Concept:

The equation of ellipse is \(\rm \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1\;(a>b)\)

length of Latus rectum of ellipse = \(\rm \frac{2b^2}{a}\)

Length of minor axis = 2b.

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

Calculations:

Given, the latus rectum of an ellipse is equal to half of the minor axis.

Suppose, the equation of ellipse is \(\rm \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1\)

length of Latus rectum of ellipse = \(\rm \frac{2b^2}{a}\)

Length of minor axis = 2b.

Given, the latus rectum of an ellipse is equal to half of the minor axis

⇒ \(\rm \frac{2b^2}{a} = b\)

⇒ \(\rm \frac{b}{a} = \frac{1}{2}\)

If the equation of ellipse is \(\rm \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1\) then eccentricity = \(e = \sqrt {1 - (\frac{b}{a})^2}\)

⇒ \(e = \sqrt {1 - (\frac{1}{2})^2}\)

⇒ e = \(\rm\frac {\sqrt {3}}{2}\)

If the latus rectum of an ellipse is equal to half of the minor axis, then what is its eccentricity e = \(\rm\frac {\sqrt {3}}{2}\)

Calculation

be = 13, b = 5

a² = b² (e² – 1)

= b² e² – b²

= 169 – 25 = 144

\(\ell(\mathrm{LR})=\frac{2 \mathrm{a}^{2}}{\mathrm{~b}}=\frac{2 \times 144}{5}=\frac{288}{5}\)

Hence option 3 is correct