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JEE MATHEMATICS MCQ: PARABOLA

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Find length of focal chord of parabola y2=8xy^2=8 x which is perpendicular to line x+y=1x+y=1
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Correct option: B
828 \sqrt{2}
Option B (828 \sqrt{2}) is correct because the length of the focal chord perpendicular to the given line is calculated using the formula 4acsc⁡2(θ)4a \csc^2(\theta), where θ\theta is the angle the chord makes with the x-axis.

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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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