AP EAMCET202017 Sep 2020Evening ShiftMathematicsParabolaActual
The common tangent to the parabola (y^2=32 x ) and (x^2=256 y ) will be
Options
- A(2 x+4 y+64=0 )
- B(x+2 y-32=0 )
- C(2 x+4 y+32=0 )
- D(4 x+2 y+64=0 )
Correct answer
A. (2 x+4 y+64=0 )
Step-by-step solution
Given parabolas are, (y^2=32 x and x^2=256 y ) We use a standard result to find equation of common tangent. Equation of tangent common to (y^2=4 a x ) and (x^2=4 b y ) is, (b^ 1 / 3 y+a^ 1 / 3 x+ (a^2 b^2 )^ 1 / 3 =0 ) Here, ( aligned & 4 a=32 a=8 & 4 b=256, b=64 aligned ) So tangent is, ( array ll & (64)^ 1 / 3 y+(8)^ 1 / 3 x+ (8^2 64^2 )^ 1 / 3 =0 & 4 y+2 x+64=0 array )