Hyperbola is an important topic in the Mathematics section in VITEEE exam. Practising this topic will increase your score overall and make your conceptual grip on VITEEE exam stronger.
This article gives you a full set of VITEEE PYQs for Hyperbola with explanations for effective preparation. Practice of VITEEE Mathematics PYQs including Hyperbola questions regularly will improve accuracy, speed, and confidence in the VITEEE 2026 exam.
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VITEEE PYQs for Hyperbola with Solutions
1.
The length of the straight line $x - 3y = 1$ intercepted by the hyperbola $x^2 - 4y^2 = 1$ is- $\sqrt{10} $
- $\frac{6}{5}$
- $ \frac{1}{\sqrt{10}} $
- $\frac{6}{5} \sqrt{10} $
2.
The equation of the hyperbola with vertices (3,0), (-3,0) and semi-latus rectum 4 is given by:- $4x^2 - 3y^2 + 36 = 0$
- $4x^2 - 3y^2 + 12 = 0$
- $4x^2 - 3y^2 - 36 = 0$
- $4x^2 -+3y^2 - 25 = 0$
3.
If the eccentricity of the hyperbola $x^{2}-y^{2} cos ec^{2} \alpha=25 is \sqrt{5}$ times the eccentricity of the ellipse $x^{2} cos ec^{2} \alpha+y^{2}=5, then \alpha$ is equal to :- $tan^{-1} \sqrt{2}$
- $ sin^{-1} \sqrt{\frac{3}{4}}$
- $ tan^{-1} \sqrt{\frac{2}{5}}$
- $ sin^{-1}\sqrt{\frac{2}{5}}$
4.
The combined equation of the asymptotes of the hyperbola $2x^2 + 5xy + 2y^2 + 4x + 5y = 0$ is -- $2x^2 + 5 xy + 2y^2 + 4x + 5y + 2 = 0$
- $2x^2 + 5 xy + 2y^2 + 4x + 5y - 2 = 0$
- $2x^2 + 5 xy + 2y^2 = 0$
- None of these
5.
The line $2x+\sqrt{6y}=2$ is a tangent to the curve $x^{2}-2y^{2}=4$. The point of contact is- $\left(4,-\sqrt{6}\right)$
- $\left(7,-2\sqrt{6}\right)$
- $\left(2,3\right)$
- $\left(\sqrt{6},1\right)$
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