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For the ellipse 4x^(2)+5y^(2)-16x-30y+60...

For the ellipse `4x^(2)+5y^(2)-16x-30y+60=0`, which of the following is true ?

A

Centre=(2,3)

B

Length of major axes is1

C

Eccentricity=`(1)/(sqrt5)`

D

All of these

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the given equation of the ellipse and determine its properties, such as the center, lengths of the axes, and eccentricity. ### Step-by-Step Solution: 1. **Rewrite the given equation**: The equation of the ellipse is given as: \[ 4x^2 + 5y^2 - 16x - 30y + 60 = 0 \] 2. **Rearranging the equation**: Move the constant term to the right side: \[ 4x^2 + 5y^2 - 16x - 30y = -60 \] 3. **Completing the square for x**: For the \(x\) terms: \[ 4(x^2 - 4x) = 4((x - 2)^2 - 4) = 4(x - 2)^2 - 16 \] Substitute this back into the equation: \[ 4(x - 2)^2 - 16 + 5y^2 - 30y = -60 \] Simplifying gives: \[ 4(x - 2)^2 + 5y^2 - 30y = -44 \] 4. **Completing the square for y**: For the \(y\) terms: \[ 5(y^2 - 6y) = 5((y - 3)^2 - 9) = 5(y - 3)^2 - 45 \] Substitute this back into the equation: \[ 4(x - 2)^2 + 5(y - 3)^2 - 45 = -44 \] Simplifying gives: \[ 4(x - 2)^2 + 5(y - 3)^2 = 1 \] 5. **Dividing by 1**: The equation can be rewritten in standard form: \[ \frac{(x - 2)^2}{\frac{1}{4}} + \frac{(y - 3)^2}{\frac{1}{5}} = 1 \] 6. **Identifying the center**: From the standard form, we can see that the center of the ellipse is at: \[ (h, k) = (2, 3) \] 7. **Identifying the lengths of the axes**: The semi-major axis \(a\) and semi-minor axis \(b\) can be identified: \[ a = \frac{1}{2}, \quad b = \frac{1}{\sqrt{5}} \] Therefore, the lengths of the axes are: - Length of the major axis = \(2b = 2 \times \frac{1}{\sqrt{5}} = \frac{2}{\sqrt{5}}\) - Length of the minor axis = \(2a = 2 \times \frac{1}{2} = 1\) 8. **Calculating eccentricity**: The eccentricity \(e\) is given by the formula: \[ e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{\left(\frac{1}{\sqrt{5}}\right)^2}{\left(\frac{1}{2}\right)^2}} = \sqrt{1 - \frac{\frac{1}{5}}{\frac{1}{4}}} = \sqrt{1 - \frac{4}{5}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}} \] ### Summary of Results: - Center: \((2, 3)\) - Length of major axis: \(\frac{2}{\sqrt{5}}\) - Length of minor axis: \(1\) - Eccentricity: \(\frac{1}{\sqrt{5}}\) ### Conclusion: All the options provided in the question regarding the properties of the ellipse are correct.
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