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JEE MATHEMATICS MCQ: PARABOLA

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Two common tangents to the circle x2+y2=2a2x^2+y^2=2 a^2 and parabola y2=8axy^2=8ax are
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Correct option: B
y=±(x+2a)y=\pm(x+2 a)
Option B (y=±(x+2a)y=\pm(x+2 a)) is correct because by finding the tangent to the parabola and then applying the condition for tangency to the circle, we arrive at this equation.

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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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