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If one vertex of an ellipse is (0,7) and...

If one vertex of an ellipse is (0,7) and the corresponding directrix is y = 12 , then the equation of the ellipse is

A

`95x^(2) + 144y^(2) = 4655`

B

`144x^(2)+95y^(2) = 4655`

C

`95x^(2) + 144y^(2) = 13680`

D

none of these

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The correct Answer is:
To find the equation of the ellipse given one vertex at (0, 7) and the corresponding directrix at y = 12, we can follow these steps: ### Step 1: Identify the values of \( a \) and \( e \) The vertex of the ellipse is given as (0, 7). In the standard form of an ellipse, the vertex is located at (0, a). Thus, we have: - \( a = 7 \) The directrix is given by the equation \( y = 12 \). The relationship between the vertex, directrix, and the eccentricity \( e \) is given by: \[ \frac{a}{e} = 12 \] Substituting the value of \( a \): \[ \frac{7}{e} = 12 \] From this, we can solve for \( e \): \[ e = \frac{7}{12} \] ### Step 2: Find the value of \( c \) The eccentricity \( e \) is also defined as: \[ e = \frac{c}{a} \] Substituting the known values: \[ \frac{c}{7} = \frac{7}{12} \] Cross-multiplying gives: \[ c = \frac{7 \times 7}{12} = \frac{49}{12} \] ### Step 3: Use the relationship \( a^2 = b^2 + c^2 \) We know: - \( a^2 = 7^2 = 49 \) - \( c^2 = \left(\frac{49}{12}\right)^2 = \frac{2401}{144} \) Using the relationship: \[ a^2 = b^2 + c^2 \] Substituting the known values: \[ 49 = b^2 + \frac{2401}{144} \] To solve for \( b^2 \), first convert 49 into a fraction with a denominator of 144: \[ 49 = \frac{49 \times 144}{144} = \frac{7056}{144} \] Now, substituting: \[ \frac{7056}{144} = b^2 + \frac{2401}{144} \] Subtracting \( \frac{2401}{144} \) from both sides: \[ b^2 = \frac{7056 - 2401}{144} = \frac{4655}{144} \] ### Step 4: Write the equation of the ellipse Since the major axis is along the y-axis, the standard form of the ellipse is: \[ \frac{y^2}{a^2} + \frac{x^2}{b^2} = 1 \] Substituting \( a^2 \) and \( b^2 \): \[ \frac{y^2}{49} + \frac{x^2}{\frac{4655}{144}} = 1 \] To eliminate the fraction in \( b^2 \), multiply through by \( 144 \): \[ \frac{144y^2}{49} + \frac{144x^2}{\frac{4655}{144}} = 144 \] This simplifies to: \[ 144y^2 + \frac{144 \cdot 144 x^2}{4655} = 144 \cdot 49 \] Thus, the equation of the ellipse can be expressed as: \[ 95y^2 + 144x^2 = 4655 \] ### Final Equation The final equation of the ellipse is: \[ 95y^2 + 144x^2 = 4655 \]
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  10. In the ellipse 9x^(2) + 5y^(2) = 45 , distance between the foci is

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  11. If the distance foci of an ellipse is 8 and distance between its dire...

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  12. Find the equation of the hyperbola whole transverse and conjugate axes...

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  13. Find the equation of hyperbola where, conjugate axis is 3 along Y-axis...

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  14. The length of the transverse axis of a hyperbola is 7 and it passes th...

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  15. One focus at (4,0) corresponding directrix x = 1 .

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  17. Conjugate axis = 10 and eccentricity = 6/5

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  18. One focus at (3,0) and eccentricity = 6/5

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  19. Find the equation of hyperbola whose conjugate axis = latus-rectum = 8...

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