JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Hyperbola

For some

θ \in (0, \frac{π}{2})

, let the eccentricity and the length of the latus rectum of the hyperbola

x^{2} - y^{2} sec^{2} θ = 8

e_{1}

and

ℓ_{1}

, respectively, and let the eccentricity and the length of the latus rectum of the ellipse

x^{2} sec^{2} θ + y^{2} = 6

e_{2}

and

ℓ_{2}

, respectively. If

e_{1}^{2} = e_{2}^{2} (sec^{2} θ + 1)

, then

(\frac{ℓ _{1} ℓ _{2}}{e _{1} e _{2}}) tan^{2} θ

is equal to ______.

Answer: 8

Solution

The hyperbola

\frac{x ^{2}}{8} - \frac{y ^{2}}{8 c o s ^{2} θ} = 1

has

e_{1}^{2} = 1 + cos^{2} θ

and

ℓ_{1} = \frac{2 \cdot 8 c o s ^{2} θ}{8} = 42 cos^{2} θ

. The ellipse

\frac{x ^{2}}{6 c o s ^{2} θ} + \frac{y ^{2}}{6} = 1

has its major axis along

y

, so

e_{2}^{2} = 1 - cos^{2} θ = sin^{2} θ

and

ℓ_{2} = \frac{2 \cdot 6 c o s ^{2} θ}{6} = 26 cos^{2} θ

. The relation

e_{1}^{2} = e_{2}^{2} (sec^{2} θ + 1)

becomes

1 + cos^{2} θ = sin^{2} θ (sec^{2} θ + 1)

, i.e.

2 cos^{4} θ = 1 - cos^{2} θ

, giving

cos^{2} θ = \frac{1}{2}

and

θ = \frac{π}{4}

. Then

e_{1} = \frac{3}{2}, ℓ_{1} = 22, e_{2} = \frac{1}{2}, ℓ_{2} = 6

and

tan^{2} θ = 1

, so

(\frac{ℓ _{1} ℓ _{2}}{e _{1} e _{2}}) tan^{2} θ = \frac{2 2 \cdot 6}{3/2 \cdot ( 1/ 2 )} = 8

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Integer answer