Coordinate Geometry – Conic Sections (Ellipse & Hyperbola) – JEE Main 28 Jan 2026 Shift 2

Solution:

For the Ellipse $E$:
Equation: $\frac{x^2}{144} + \frac{y^2}{169} = 1$.
Since $169 > 144$, the major axis is along the y-axis.
$a^2 = 144, b^2 = 169$. (Note: usually $a > b$, but here let’s stick to $x^2/A^2 + y^2/B^2$ where $B>A$).
Eccentricity $e_E = \sqrt{1 – \frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13}$.
Foci are at $(0, \pm be_E) = (0, \pm 13 \times \frac{5}{13}) = (0, \pm 5)$.

For the Hyperbola $H$:
Equation: $\frac{x^2}{16} – \frac{y^2}{\lambda^2} = -1 \Rightarrow \frac{y^2}{\lambda^2} – \frac{x^2}{16} = 1$.
This is a vertical hyperbola.
Semi-transverse axis $A = \lambda$, Semi-conjugate axis $B = 4$ ($B^2=16$).
Eccentricity $e = \sqrt{1 + \frac{B^2}{A^2}} = \sqrt{1 + \frac{16}{\lambda^2}}$.
Foci are at $(0, \pm Ae) = (0, \pm \lambda e)$.

Since the foci are the same:
$$\lambda e = 5$$
Squaring both sides:
$$\lambda^2 e^2 = 25$$
$$\lambda^2 (1 + \frac{16}{\lambda^2}) = 25$$
$$\lambda^2 + 16 = 25 \Rightarrow \lambda^2 = 9 \Rightarrow \lambda = 3$$

Now we calculate $e$ and $L$ for the hyperbola.
$$e = \sqrt{1 + \frac{16}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3}$$
Latus Rectum $L = \frac{2(\text{conjugate})^2}{\text{transverse}} = \frac{2(16)}{3} = \frac{32}{3}$.

Value to find: $24(e + L)$
$$24(e + L) = 24(\frac{5}{3} + \frac{32}{3}) = 24(\frac{37}{3})$$
$$= 8 \times 37 = 296$$

Ans. (1)