A tangent is drawn to parabola y2−4x+4=0 at a point P which cuts the directrix at the point Q. If a point R is such that it divides QP externally in ratio 1:2, then the locus of point R is
A
y(x+1)2+4=0
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B
y2(x+1)+4=0
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C
y=0
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D
(x+1)(y−1)2−4=0
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Solution
The correct option is By2(x+1)+4=0
y2−4x+4=0⇒y2=4(x−1) Therefore, we get vertex : (1,0), Focus : (2,0) and Directrix : x=0 Let P be (1+t2,2t) ∴ Slope of the tangent =1t Equation of tangent at P is y−2t=1t(x−(1+t2))⇒y−2t=xt−1+t2t
Q lies on the directrix x=0 ∴ Coordinates of Q is (0,t−1t) and P(1+t2,2t) ∵R divides PQ in 1:2 externally ∴R(x,y)=⎛⎜
⎜
⎜
⎜⎝1+t2−1,2t−2(t−1t)−1⎞⎟
⎟
⎟
⎟⎠