If the parametric equations of the parabola are given by x  4t2  2t 1; 2 y  3t  t 1
and the vertex of the parabola also satisfies
k
y x ,
100
  then the area of the circle
2 2 x  y 12x 10y  2k  0 in square units is

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Saurabh Koranglekar · Answer

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If the parametric equations of the parabola are given by x  4t2  2t 1; 2 y  3t  t 1 and the vertex of the parabola also satisfies k y x , 100   then the area of the circle 2 2 x  y 12x 10y  2k  0 in square units is

If the parametric equations of the parabola are given by x  4t2  2t 1; 2 y  3t  t 1
and the vertex of the parabola also satisfies
k
y x ,
100
  then the area of the circle
2 2 x  y 12x 10y  2k  0 in square units is

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If the parametric equations of the parabola are given by x  4t2  2t 1; 2 y  3t  t 1 and the vertex of the parabola also satisfies k y x , 100   then the area of the circle 2 2 x  y 12x 10y  2k  0 in square units is

If the parametric equations of the parabola are given by x  4t2  2t 1; 2 y  3t  t 1and the vertex of the parabola also satisfiesky x ,100  then the area of the circle2 2 x  y 12x 10y  2k  0 in square units is

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If the parametric equations of the parabola are given by x  4t2  2t 1; 2 y  3t  t 1
and the vertex of the parabola also satisfies
k
y x ,
100
  then the area of the circle
2 2 x  y 12x 10y  2k  0 in square units is