The equation of the common tangent to the curves $y^2 = 8x$ and $xy = -1$ is

  • A
    $3y = 9x + 2$
  • B
    $y = 2x + 1$
  • C
    $2y = x + 8$
  • D
    $y = x + 2$

Explore More

Similar Questions

$A$ particle starting from the origin $(0, 0)$ moves in a straight line in the $(x, y)$ plane. Its coordinates at a later time are $(\sqrt{3}, 3)$. The path of the particle makes an angle of $.......^{\circ}$ with the $x$-axis.

Among $K$,$Mg$,$Au$,and $Cu$,the one which is extracted by heating its ore in air is

The bonds between $P$ atoms and $Cl$ atoms in $PCl_5$ are likely to be

Which of the following is a polyamide?

Consider a block of conducting material of resistivity ' $\rho$ ' as described. Current ' $I$ ' enters at ' $A$ ' and leaves from ' $D$ '. We apply the superposition principle to find the voltage ' $\Delta V$ ' developed between ' $B$ ' and ' $C$ '. The calculation is done in the following steps:
$(i)$ Take current ' $I$ ' entering from ' $A$ ' and assume it to spread over a hemispherical surface in the block.
(ii) Calculate the field $E(r)$ at distance ' $r$ ' from $A$ by using Ohm's law $E = \rho j$,where $j$ is the current per unit area at ' $r$ '.
(iii) From the ' $r$ ' dependence of $E(r)$,obtain the potential $V(r)$ at $r$.
(iv) Repeat $(i)$,(ii),and (iii) for current ' $I$ ' leaving ' $D$ ' and superpose results for ' $A$ ' and ' $D$ '.
For current entering at $A$,the electric field at a distance ' $r$ ' from $A$ 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