JEE Mathematics 2023 Question with Solution

Given: The ellipse has centre $(1,0)$, major axis along the $x$-axis, latus rectum length $\frac{1}{2}$, and the minor axis subtends angle $$ 60^\circ

$$$$

Find: The square of the sum of the lengths of the minor and major axes.

Take the standard form

$$\frac{(x-1)^2}{a^2}+\frac{y^2}{b^2}=1$$

where major axis length is $2a$ and minor axis length is $$ 2b.

$$$$

Using the latus rectum condition,

$$\frac{2b^2}{a}=\frac{1}{2}$$

so

$$b^2=\frac{a}{4}.$$

Let the distance between the foci be $$ 2c

$$$$

c^2=a^2-b^2.

Since the minor axis subtends angle $$ 60^\circ $$ at the foci, using the stated relation,

\sin\frac{\theta}{2}=\frac{b}{c}

with $$ \theta=60^\circ $$ gives

\sin 30^\circ=\frac{b}{c}

$$so$$

\frac{1}{2}=\frac{b}{c}

$$and hence$$

c=2b.

$$$$

Now substitute into

$$c^2=a^2-b^2:$$ $$(2b)^2=a^2-b^2$$ $$4b^2=a^2-b^2$$ $$a^2=5b^2.$$

Thus

$$a=\sqrt{5}\,b.$$

From $b^2=\frac{a}{4}$, substitute $$ a=\sqrt{5},b

$$$$

b^2=\frac{\sqrt{5},b}{4}

$$which gives$$

b=\frac{\sqrt{5}}{4}.

$$Therefore,$$

a=\sqrt{5}\cdot \frac{\sqrt{5}}{4}=\frac{5}{4}.

$$$$

Now,

$$2a+2b=2\left(\frac{5}{4}+\frac{\sqrt{5}}{4}\right)=\frac{5+\sqrt{5}}{2}.$$

So,

$$(2a+2b)^2=\left(\frac{5+\sqrt{5}}{2}\right)^2=\frac{30+10\sqrt{5}}{4}.$$

the solution concludes with Final Answer: $9$. Therefore, taking the solution, the answer is $9$.

The algebra shown in the solution leads to

$$(2a+2b)^2=\left(\frac{5+\sqrt{5}}{2}\right)^2=\frac{15+5\sqrt{5}}{2},$$

which is not equal to $$ 9.

However, the solution explicitly states **Final Answer: $$9$$**. As instructed, the final stated answer on the solution is used as the authoritative answer, while noting this inconsistency.