A hyperbola passes through the foci of the ellipse <math><mrow><mfrac><mrow><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><mn>2</mn><mn>5</mn></mrow></mfrac><mo>+</mo><mfrac><mrow><msup><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><mn>1</mn><mn>6</mn></mrow></mfrac><mo>=</mo><mn>1</mn></mrow></math> and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is

x225+y216=116 = 25 (1 – e2)⇒e=35OF1 = 5 (35)=3For Hyperbola : e’ =53a = 3Hyperbola, x29−y216=1

A hyperbola passes through 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>2</mn><mn>5</mn></mrow></mfrac><mo>+</mo><mfrac><mrow><msup><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><mn>1</mn><mn>6</mn></mrow></mfrac><mo>=</mo><mn>1</mn></mrow></math>&nbsp;and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is

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Conic Sections Ncert Solutions Maths class 11th Maths Class 11th

A hyperbola passes through the foci of the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is

Option 1 -

$x^{2} - y^{2} = 9$

Option 2 -

$\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$

Option 3 -

$\frac{x^{2}}{9} - \frac{y^{2}}{25} = 1$

Option 4 -

$\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$

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Answered by

Payal Gupta | Contributor-Level 10

a month ago

Correct Option - 2

Detailed Solution:

$\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$

16 = 25 (1 – e²)

$\Rightarrow e = \frac{3}{5}$

OF₁= 5 $(\frac{3}{5}) = 3$

For Hyperbola : e’ $= \frac{5}{3}$

a = 3

$H y p e r b o l a, \frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$

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Or 4 (1 – e²) = e²

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

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

⇒ 2y – 6 = x – 2

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

...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

4x + 2y – 12 = 0

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

K $= \frac{4 \sqrt[]{3}}{\sqrt[]{3} x + y}$ $\frac{}{}$

K = $\frac{\sqrt[]{3} x - y}{4 \sqrt[]{3}}$ $\frac{}{}$

$∴ \frac{4 \sqrt[]{3}}{\sqrt[]{3} x + y} = \frac{\sqrt[]{3} x - y}{4 \sqrt[]{3}}$

$\frac{x^{2}}{16} - \frac{y^{2}}{48} = 1$

48 = 16 (e2 – 1)

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

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A hyperbola passes through the foci of the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is

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A hyperbola passes through the foci of the ellipse x225+y216=1 and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is

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A hyperbola passes through the foci of the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is