Latus Rectum MCQ Quiz in मल्याळम - Objective Question with Answer for Latus Rectum - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

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

The required area is in the shape of kite.

Area of kite \(=\dfrac{1}{2}\times \text{(Product of its diagonal)}\)

\(\Rightarrow \text{Eccentricity }(e)=\dfrac{3}{5}\)

\(\Rightarrow \text{Distance between foci }=6\)

\(\therefore 2ae=6\)

\(\Rightarrow ae=3\)

\(\Rightarrow a\times\dfrac{3}{5}=3\)

\(\therefore a=5\)

Now,

\(\Rightarrow b^2=a^2-(ae)^2\)

\(\Rightarrow b^2=(5)^2-(3)^2\)

\(\Rightarrow b^2=25-9\)

\(\Rightarrow b^2=16\)

\(\therefore b=4\)

\(\Rightarrow \text{Length of major axis }(AA')=2a\)

\(=2\times 5\)

\(=10\)

\(\Rightarrow \text{Length of minor axis }(BB')=2b\)

\(=2\times 4\)

\(=8\)

\(\Rightarrow \text{Required area }=\text{Area of kite }ABA'B'\)

\(=\dfrac{1}{2}\times (AA')\times (BB')\)

\(=\dfrac{1}{2}\times 10\times 8\)

\(=40\)

\(\dfrac{2b}{a}^2 = 8\)

and

\(2ae = 2b\)

\(\Longrightarrow \dfrac{b}{a} = e\ and\ 1 - e^2 = e^2 \Longrightarrow e = \dfrac{1}{\sqrt{2}}\)

\(\Longrightarrow b = 4\sqrt{2}\ and\ a + 8\)

so equation of ellipse is

\(\dfrac{x^2}{64} + \dfrac{y^2}{32} = 1\)

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:

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 )