A) \[{{y}^{2}}-4x+2=0\]
B) \[{{y}^{2}}+4x+2=0\]
C) \[{{x}^{2}}+4y+2=0\]
D) \[{{x}^{2}}-4y+2=0\]
Correct Answer: A
Solution :
Coordinates of P are (1, 0). Let the coordinates of the point Q are\[(2{{t}^{2}},4t)\]and mid-point of PQ is\[(h,k),\] Then, \[2h=2{{t}^{2}}+1\] and \[2k=4t\Rightarrow t=\frac{k}{2}\] \[\therefore \] \[2h=\frac{2{{k}^{2}}}{4}+1\] \[\Rightarrow \] \[4h={{k}^{2}}+2\] \[\therefore \]Locus of\[(h,k)\]is \[{{y}^{2}}-4x+2=0\]
You need to login to perform this action.
You will be redirected in
3 sec
