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JEE MATHEMATICS MCQ: PARABOLA

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The parabola with directrix x+2y−1=0x + 2y - 1 = 0 and focus (1,0)\left( {1,0} \right) is
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Correct option: A
4x2−4xy+y2−8x+4y+4=04{x^2} - 4xy + {y^2} - 8x + 4y + 4 = 0
Option A (4x2−4xy+y2−8x+4y+4=04{x^2} - 4xy + {y^2} - 8x + 4y + 4 = 0) is correct because it represents the equation of the parabola with the given directrix and focus.

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The equation of a common tangent to the parabolas y=x2y=x^2 and y=−(x−2)2y=-(x-2)^2 is A point PP moves such that slope of one tangent drawn from it to parabola y2=16xy^2=16 x is four times the slope of other tangent, then the length of latus rectum of the conic represented by locus of PP, is Let P,Q,RP,Q,R be three points on parabola y2=4x{y^2} = 4x and normal at PP and RR meet at QQ, then the locus of the mid-point of the chord PRPR is a parabola whose vertex is at The tangents at the end points of any chord through (1,0)(1,0) to the parabola y2+4x=8y^2+4 x=8 intersect The focus of the parabola x2=−16yx^2=-16 y is If II denotes the semi-latusrectum of the parabola y2=4ax{y^2} = 4ax and SPSP and SQSQ denote the segments of any focal chord PQ,SPQ,S being the focus, then SP,lSP,l and SQSQ are in the relation
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