The equation $\frac{x^2}{12 - \lambda} + \frac{y^2}{8 - \lambda} = 1$ represents:

  • A
    If $\lambda < 8$,then it is a hyperbola.
  • B
    If $\lambda > 0$,then it is an ellipse.
  • C
    If $8 < \lambda < 12$,then it is a hyperbola.
  • D
    None of the above.

Explore More

Similar Questions

If $P(x_1, y_1)$ is a point on the hyperbola $x^2 - y^2 = a^2$,then $SP \cdot S'P = \_\_\_\_$

Let $H: \frac{-x^2}{a^2}+\frac{y^2}{b^2}=1$ be the hyperbola,whose eccentricity is $\sqrt{3}$ and the length of the latus rectum is $4\sqrt{3}$. Suppose the point $(\alpha, 6)$,where $\alpha > 0$,lies on $H$. If $\beta$ is the product of the focal distances of the point $(\alpha, 6)$,then $\alpha^2+\beta$ is equal to:

Let $P(a \sec \theta, b \tan \theta)$ and $Q(a \sec \phi, b \tan \phi)$ where $\theta + \phi = \frac{\pi}{2}$ be two points on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. If $(h, k)$ is the point of intersection of the normals drawn at $P$ and $Q$,then $k=$

The equation $\frac{1}{r} = \frac{1}{8} + \frac{3}{8} \cos \theta$ represents:

The equation of the hyperbola which passes through the point $(2,3)$ and has the asymptotes $4x+3y-7=0$ and $x-2y-1=0$ is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo