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The equation of the common tangent of the parabola y2=32x and x2=108y is
A) x=0
B) 2x-3y-36=0
C) 2x+3y+36=0
D) 2x-3y+36=0

Dear Student ,
 
Please find below the solution to the asked query :

Equation of tangent to the parabola y2=32x isy=mx+8m  ...1    Since 4a=32 ⇒ a=8It meets x2=108ySo ,x2=108 mx+8mmx2=108m2x+864mx2-108m2 x-864=0If the line 1 is the tangent to the second parabola , then roots of above equation must be equal .So ,-108m22-4 m -864=0     Since D=b2-4ac=027m3+8=0m3=-827m=-23Substituting the value of m in 1 , we gety=-2x3+8-23y=-2x3-12y=-2x-3633y=-2x-362x+3y+36=0 , is the required tangent .
 
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