JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Hyperbola

Let the foci of hyperbola coincide with the foci of the ellipse $\frac{x^2}{36}+\frac{y^2}{16}=1$. If the eccentricity of the hyperbola is 5 , then the length of its latus rectum is:

If the line $\alpha x+2y=1$, where $\alpha \in \mathbb{R}$, does not meet the hyperbola $x^2-9y^2=9$, then a possible value of $\alpha$ is:

Let $P(10,2\sqrt{15})$ be a point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, whose foci are S and $S^{\prime }$. If the length of its latus rectum is 8, then the square of the area of $△PS{S}^{\prime }$ is equal to :

Let PQ be a chord of the hyperbola $\frac{{\mathrm{x}}^2}{4}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$, perpendicular to the x-axis such that OPQ is an equilateral triangle, $O$ being the centre of the hyperbola. If the eccentricity of the hyperbola is $\sqrt{3}$, then the a …

For some $\theta \in \left(0,\frac{\pi }{2}\right)$, let the eccentricity and the length of the latus rectum of the hyperbola $x^2-{\mathrm{y}}^2{\sec }^2\theta =8$ be ${\mathrm{e}}_1$ and ${\ell }_1$, respectively, and let the eccentricity and the length of the latus r …

Let the ellipse $\mathrm{E}:\frac{{\mathrm{x}}^2}{144}+\frac{{\mathrm{y}}^2}{169}=1$ and the hyperbola $\mathrm{H}:\frac{{\mathrm{x}}^2}{16}-\frac{{\mathrm{y}}^2}{{\lambda }^2}=-1$ have the same foci. If e and L respectively denote the eccentricity and the length of the latu …