Q95 / 160 · AP EAMCET 2024 (20 May Shift 1) · Mathematics

If a tangent to the hyperbola $x^2-\frac{y^2}{3}=1$ is also a tangent to the parabola $y^2=8 x$. then equation of such tangent with the positive slope is

  • A
    $y-x-\frac{1}{2}=0$
  • Correct answer
    B
    $y-2 x-1=0$
  • C
    $2 y-4 x-1=0$
  • D
    $y-x-1=0$

Solution

Eq. of tangent to $y^2=8 x$ is $y=m x+\frac{2}{m}$ Eq. of tangent to $x^2-\frac{y^2}{3}=1$ is $y=m x \pm \sqrt{m^2-3}$ On comparing we get $m= \pm 2$ $\Rightarrow$ tangent with + ve slope is $y-2 x-1=0$

Hence, the correct answer is option: 'B'.

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