If $\frac{x}{a} + \frac{y}{b} = 1$ is a tangent to the curve $x = Kt, y = \frac{K}{t}, K > 0$,then

  • A
    $a > 0, b > 0$
  • B
    $a > 0, b < 0$
  • C
    $a < 0, b < 0$
  • D
    Both $(A)$ and $(C)$

Explore More

Similar Questions

If $P(\theta)$ lies on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ and $S$ and $S^{\prime}$ are the foci of the hyperbola,then $SP \cdot S^{\prime}P =$

The equation of a tangent to the hyperbola $16x^2 - 25y^2 - 96x + 100y - 356 = 0$ which makes an angle $45^{\circ}$ with its transverse axis is

Let $A(2 \sec \theta, 3 \tan \theta)$ and $B(2 \sec \phi, 3 \tan \phi)$ where $\theta+\phi=\frac{\pi}{2}$ be two points on the hyperbola $\frac{x^2}{4}-\frac{y^2}{9}=1$. If $(\alpha, \beta)$ is the point of intersection of normals to the hyperbola at $A$ and $B$,then $\beta$ is equal to

Let $H: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ be a hyperbola such that the distance between its foci is $6$ and the distance between its directrices is $\frac{8}{3}$. If the line $x = k$ intersects the hyperbola $H$ at the points $A$ and $B$ such that the area of the triangle $AOB$ is $4\sqrt{15}$,where $O$ is the origin,then $a^2$ equals

Consider a branch of the hyperbola $x^2 - 2y^2 - 2\sqrt{2}x - 4\sqrt{2}y - 6 = 0$ with vertex at the point $A$. Let $B$ be one of the end points of its latus rectum. If $C$ is the focus of the hyperbola nearest to the point $A$,then the area of the triangle $ABC$ 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