Question

Let an ellipse with centre (1, 0) and latus rectum of length \(\frac{1}{2}\) have its major axis along x-axis. If its minor axis subtends an angle 60° at the foci, then the square of the sum of the lengths of its minor and major axes is equal to

Answer 1

Correct Answer: 9

To solve the problem, we need to find the square of the sum of the lengths of the minor and major axes of the ellipse given the conditions in the question.
Step 1: Identify the parameters of the ellipse
The center of the ellipse is given as (1, 0), and since the major axis is along the x-axis, the standard form of the ellipse can be written as:
$(x-1)^2/a^2 + y^2/b^2 = 1$
where $2a$ is the length of the major axis and $2b$ is the length of the minor axis.
Step 2: Use the latus rectum length
The length of the latus rectum (LR) of the ellipse is given as 12. The formula for the latus rectum in terms of a and b is:
$LR = 2b^2/a$
Setting this equal to 12:
$2b^2/a = 12$
Multiplying both sides by a:
$2b^2 = 12a$
Dividing by 2:
$b^2 = 6a$
Thus, we can express this as:
$12b^2=6a$ or $2b^2=12a$ or $b^2=6a$
(Equation 1)
Step 3: Use the angle subtended by the minor axis at the foci
The minor axis subtends an angle of 60° at the foci. The foci of the ellipse are located at $(1 \pm ae, 0)$, where $e = \sqrt{1 - b^2/a^2}$.
Using the tangent of half the angle subtended:
$tan(30°) = b/(ae)$
Since $tan(30°) = 1/\sqrt{3}$, we have:
$1/\sqrt{3} = b/(ae)$
Rearranging gives:
$b = ae/\sqrt{3}$
(Equation 2)
Step 4: Substitute e in terms of a and b
From the definition of eccentricity, we have:
$e = \sqrt{1 - b^2/a^2}$
Substituting b from Equation 2 into this gives:
$e = \sqrt{1 - (ae/\sqrt{3})^2/a^2} = \sqrt{1 - a^2e^2/(3a^2)} = \sqrt{1 - e^2/3}$
Squaring both sides:
$e^2 = 1 - e^2/3$
Multiplying through by 3:
$3e^2 = 3 - e^2$
Combining terms gives:
$4e^2 = 3 \Rightarrow e^2 = 3/4$
Step 5: Substitute back to find $b^2$
Now substituting $e^2$ back into Equation 1:
$b^2=6a$ and $b=ae/\sqrt{3}$ so $b^2=a^2e^2/3$
$6a=a^2e^2/3$
$18a=a^2e^2$
$18a=a^2(3/4)$
$18=a(3/4)$
$a=18(4/3)=24$
$b^2=6a=6(24)=144$
Step 6: Calculate the lengths of the axes
The lengths of the major and minor axes are:
$2a = 2(24) = 48$, $2b = 2\sqrt{144} = 2(12) = 24$
Step 7: Find the square of the sum of the lengths of the axes
Now we calculate:
$2a + 2b = 48 + 24 = 72$
Now squaring this sum:
$(2a + 2b)^2 = (72)^2 = 5184$
Thus, the square of the sum of the lengths of the minor and major axes is:
5184