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What is the sum of the major and minor axes of the ellipse whose eccentricity is 4/5 and length of latus rectum is 14.4 unit?
- a)32 units
- b)48 units
- c)64 units
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
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What is the sum of the major and minor axes of the ellipse whose eccen...
Let 2a and 2b be the length of major and minor axis respectively.
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What is the sum of the major and minor axes of the ellipse whose eccen...
Given:
Eccentricity (e) = 4/5
Length of latus rectum (l) = 14.4 units
We know that the eccentricity (e) of an ellipse is given by the formula:
e = c/a
where c is the distance from the center of the ellipse to one of its foci, and a is the semi-major axis.
We also know that the length of the latus rectum (l) is given by the formula:
l = 2b^2/a
where b is the semi-minor axis.
To find the sum of the major and minor axes, we need to find the values of a and b.
Solving for a in the eccentricity formula:
e = c/a
4/5 = c/a
c = 4a/5
Substituting this value of c in the latus rectum formula:
l = 2b^2/a
14.4 = 2b^2/(4a/5)
14.4 = 10b^2/a
b^2 = (14.4a)/10
b = sqrt((14.4a)/10)
Now, we can substitute the value of b in the eccentricity formula to solve for a:
e = c/a
4/5 = (4a/5)/a
4/5 = 4/5
This equation is true for any value of a.
So, we can choose any value for a. Let's assume a = 5.
Substituting the value of a in the equation for b:
b = sqrt((14.4*5)/10)
b = sqrt(7.2)
b ≈ 2.683
Therefore, the semi-major axis (a) is 5 units and the semi-minor axis (b) is approximately 2.683 units.
Sum of major and minor axes:
= 2a + 2b
= 2(5) + 2(2.683)
= 10 + 5.366
= 15.366 units
Rounded to the nearest whole number, the sum of the major and minor axes is approximately 15 units.
Hence, the correct answer is option C.
Eccentricity (e) = 4/5
Length of latus rectum (l) = 14.4 units
We know that the eccentricity (e) of an ellipse is given by the formula:
e = c/a
where c is the distance from the center of the ellipse to one of its foci, and a is the semi-major axis.
We also know that the length of the latus rectum (l) is given by the formula:
l = 2b^2/a
where b is the semi-minor axis.
To find the sum of the major and minor axes, we need to find the values of a and b.
Solving for a in the eccentricity formula:
e = c/a
4/5 = c/a
c = 4a/5
Substituting this value of c in the latus rectum formula:
l = 2b^2/a
14.4 = 2b^2/(4a/5)
14.4 = 10b^2/a
b^2 = (14.4a)/10
b = sqrt((14.4a)/10)
Now, we can substitute the value of b in the eccentricity formula to solve for a:
e = c/a
4/5 = (4a/5)/a
4/5 = 4/5
This equation is true for any value of a.
So, we can choose any value for a. Let's assume a = 5.
Substituting the value of a in the equation for b:
b = sqrt((14.4*5)/10)
b = sqrt(7.2)
b ≈ 2.683
Therefore, the semi-major axis (a) is 5 units and the semi-minor axis (b) is approximately 2.683 units.
Sum of major and minor axes:
= 2a + 2b
= 2(5) + 2(2.683)
= 10 + 5.366
= 15.366 units
Rounded to the nearest whole number, the sum of the major and minor axes is approximately 15 units.
Hence, the correct answer is option C.
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What is the sum of the major and minor axes of the ellipse whose eccentricity is 4/5 and length of latus rectum is 14.4 unit?a)32 unitsb)48 unitsc)64 unitsd)None of theseCorrect answer is option 'C'. Can you explain this answer?
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