Trigonometry
Trigonometric Ratio and Identites
NumericalMCQ (Single Correct Answer)
1
JEE Main 2025 (Online) 7th April Morning Shift
Numerical
+4
-1

Consider the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ having one of its focus at $\mathrm{P}(-3,0)$. If the latus ractum through its other focus subtends a right angle at P and $a^2 b^2=\alpha \sqrt{2}-\beta, \alpha, \beta \in \mathbb{N}$, then $\alpha+\beta$ is _________ .

Your input ____
2
JEE Main 2025 (Online) 3rd April Evening Shift
Numerical
+4
-1
If the equation of the hyperbola with foci $(4,2)$ and $(8,2)$ is $3 x^2-y^2-\alpha x+\beta y+\gamma=0$, then $\alpha+\beta+\gamma$ is equal to__________.
Your input ____
3
JEE Main 2025 (Online) 3rd April Morning Shift
Numerical
+4
-1

Let the product of the focal distances of the point $\mathbf{P}(4,2 \sqrt{3})$ on the hyperbola $\mathrm{H}: \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ be 32 . Let the length of the conjugate axis of H be $p$ and the length of its latus rectum be $q$. Then $p^2+q^2$ is equal to__________

Your input ____
4
JEE Main 2025 (Online) 24th January Evening Shift
Numerical
+4
-1

Let $\mathrm{H}_1: \frac{x^2}{\mathrm{a}^2}-\frac{y^2}{\mathrm{~b}^2}=1$ and $\mathrm{H}_2:-\frac{x^2}{\mathrm{~A}^2}+\frac{y^2}{\mathrm{~B}^2}=1$ be two hyperbolas having length of latus rectums $15 \sqrt{2}$ and $12 \sqrt{5}$ respectively. Let their ecentricities be $e_1=\sqrt{\frac{5}{2}}$ and $e_2$ respectively. If the product of the lengths of their transverse axes is $100 \sqrt{10}$, then $25 \mathrm{e}_2^2$ is equal to _________ .

Your input ____
JEE Main Subjects