Question

Find the equation of the common tangents to the parabola $y^2=8x$ and hyperbola $3x^2-y^2=3.$

• A. $2x-y+1=0$ and $2x+y+1=0.$
• B. $2x-y+1=0$ and $2x+y-1=0.$
• C. $-2x+y+1=0$ and $2x+y+1=0.$
• D. None of these
  

Answer 1

The correct option is A $2x-y+1=0$ and $2x+y+1=0.$

$y^2=8x\therefore 4a=8$ or $a=2$
Any tangent to the parabola is
$y=mx+2/m$ ... (1)
If it is also a tangent to the hyperbola
$x^2/1-y^2/3=1i.e.a^2=1,b^2=3,$ then
$c^2=a^2{m}^2-b^2$ or $4/m^2=1.m^2-3.$
or $m^4-3m^2-4=0$ or $(m^2-4)(m^2+1)=0$
$\therefore m=\pm 2.$ Putting for min (1). we get the tangents as
$2x-y+1=0$ and $2x+y+1=0.$