Let the foci of the ellipse <math><mrow><mfrac><mrow><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><mn>1</mn><mn>6</mn></mrow></mfrac><mo>+</mo><mfrac><mrow><msup><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><mn>7</mn></mrow></mfrac><mo>=</mo><mn>1</mn></mrow></math> and the hyperbola <math><mrow><mfrac><mrow><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><mn>1</mn><mn>4</mn><mn>4</mn></mrow></mfrac><mo>−</mo><mfrac><mrow><msup><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><mi>α</mi></mrow></mfrac><mo>=</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>2</mn><mn>5</mn></mrow></mfrac></mrow></math> coincide. Then the length of the latus rectum of the hyperbola is

Ellipse : x216+y27=1Eccentricity = 1−710=34Foci ≡ (±ae, 0) ≡ (±3, 0)Hyperbola : x2 (14425)−y2 (α25)=1Eccentricity = 1+α144=112144+α=2.8125125=2710

Let the foci of the ellipse&nbsp;<math><mrow><mfrac><mrow><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><mn>1</mn><mn>6</mn></mrow></mfrac><mo>+</mo><mfrac><mrow><msup><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><mn>7</mn></mrow></mfrac><mo>=</mo><mn>1</mn></mrow></math>&nbsp;and the hyperbola&nbsp;<math><mrow><mfrac><mrow><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><mn>1</mn><mn>4</mn><mn>4</mn></mrow></mfrac><mo>−</mo><mfrac><mrow><msup><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><mi>α</mi></mrow></mfrac><mo>=</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>2</mn><mn>5</mn></mrow></mfrac></mrow></math>&nbsp;coincide. Then the length of the latus rectum of the hyperbola is

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Maths NCERT Exemplar Solutions Class 12th Chapter Two Conic Sections Ncert Solutions Maths class 11th Maths Class 11th

Let the foci of the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{7} = 1$ and the hyperbola $\frac{x^{2}}{144} - \frac{y^{2}}{α} = \frac{1}{25}$ coincide. Then the length of the latus rectum of the hyperbola is

Option 1 -

$\frac{32}{9}$

Option 2 -

$\frac{18}{5}$

Option 3 -

$\frac{27}{4}$

Option 4 -

$\frac{27}{10}$

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1 Answer

Answered by

Payal Gupta | Contributor-Level 10

2 months ago

Correct Option - 4

Detailed Solution:

Ellipse : $\frac{x^{2}}{16} + \frac{y^{2}}{7} = 1$

Eccentricity = $\sqrt[]{1 - \frac{7}{10}} = \frac{3}{4}$

Foci $\equiv (\pm a e, 0)$ $\equiv (\pm 3, 0)$

Hyperbola : $\frac{x^{2}}{(\frac{144}{25})} - \frac{y^{2}}{(\frac{α}{25})} = 1$

Eccentricity = $\sqrt[]{1 + \frac{α}{144}} = \frac{1}{12} \sqrt[]{144 + α}$

$= 2 . \frac{\frac{81}{25}}{\frac{12}{5}} = \frac{27}{10}$

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...more

Slope of axis = $\frac{1}{2}$

$y - 3 = \frac{1}{2} (x - 2)$

⇒ 2y – 6 = x – 2

⇒ 2y – x – 4 = 0

2x + y – 6 = 0

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α + 1.6 = 4 ⇒ α = 2.4

β + 2.8 = 6 ⇒ β = 3.2

Ellipse passes through (2.4, 3.2)

⇒ $\frac{{(\frac{24}{10})}^{2}}{a^{2}} + \frac{{(\frac{32}{10})}^{2}}{b^{2}} = 1$

Also $1 - \frac{a^{2}}{b^{2}} = \frac{1}{2}$

$\frac{a^{2}}{b^{2}} = \frac{1}{2}$

$\frac{144}{25} b^{2} + \frac{256}{25} a^{2} = a^{2} b^{2}$

$\frac{144}{25} + \frac{256}{25} \times \frac{1}{2} = a^{2}$

$\frac{(128 + 144)}{25} = a^{2} \Rightarrow \frac{272}{25} = a^{2}$

-> $b^{2} = \frac{2 \times 272}{25}$

Latus rectum = $\frac{2 b^{2}}{a}$

(Latus rectum)²

$= \frac{4 b^{4}}{a^{2}} = 4 (\frac{b^{2}}{a^{2}}) b^{2} = \frac{8 \times 272 \times 2}{25} = \frac{4352}{25}$

less

The locus of the point of intersection of the lines $(\sqrt[]{3}) k x + k y - 4 \sqrt[]{3} = 0 a n d \sqrt[]{3} x - y - 4 (\sqrt[]{3}) k$ = 0 is a conic, whose eccentricity is…………………

Vishal Baghel

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48 = 16 (e2 – 1)

$e = \sqrt[]{4} = 2$

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Let the foci of the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{7} = 1$ and the hyperbola $\frac{x^{2}}{144} - \frac{y^{2}}{α} = \frac{1}{25}$ coincide. Then the length of the latus rectum of the hyperbola is

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Let the foci of the ellipse x216+y27=1 and the hyperbola x2144−y2α=125 coincide. Then the length of the latus rectum of the hyperbola is

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Let the foci of the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{7} = 1$ and the hyperbola $\frac{x^{2}}{144} - \frac{y^{2}}{α} = \frac{1}{25}$ coincide. Then the length of the latus rectum of the hyperbola is