If the point $(1,1)$ and the origin lie in the same region with respect to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{1} = 1$ $(a > 0)$,then the range of $a$ is

  • A
    $\left(\frac{1}{\sqrt{2}}, \infty\right)$
  • B
    $\left(0, \frac{1}{\sqrt{2}}\right)$
  • C
    $(0, 1)$
  • D
    $(0, \sqrt{2})$

Explore More

Similar Questions

If the circle $x^2+y^2=a^2$ intersects the hyperbola $xy=c^2$ in four points $(x_i, y_i)$,for $i=1, 2, 3, 4$,then $y_1+y_2+y_3+y_4$ equals

If the foci of a hyperbola are the same as that of the ellipse $\frac{x^2}{9}+\frac{y^2}{25}=1$ and the eccentricity of the hyperbola is $\frac{15}{8}$ times the eccentricity of the ellipse,then the smaller focal distance of the point $\left(\sqrt{2}, \frac{14}{3} \sqrt{\frac{2}{5}}\right)$ on the hyperbola is equal to

Find the equation of the hyperbola satisfying the given conditions: Vertices $(\pm 2, 0)$,foci $(\pm 3, 0)$.

The equation of the hyperbola whose eccentricity is $\frac{5}{3}$ and the distance between the foci is $10$ units is:

The line $y=x$ intersects the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{25}=1$ at the points $P$ and $Q$. The eccentricity of the ellipse with $PQ$ as the major axis and a minor axis of length $\frac{5}{\sqrt{2}}$ 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