1

BITSAT 2022

MCQ (Single Correct Answer)

+3

-1

A normal is drawn at the point P to the parabola $${y^2} = 8x$$, which is inclined at 60$$^\circ$$ with the straight line $$y = 8$$. Then the point P lies on the straight line

2

BITSAT 2022

MCQ (Single Correct Answer)

+3

-1

For each parabola y = x^{2} + px + q, meeting coordinate axes at 3-distinct points, if circles are drawn through these points, then the family of circles must pass through

3

BITSAT 2021

MCQ (Single Correct Answer)

+3

-1

The origin is shifted to (1, 2). The equation y^{2} $$-$$ 8x $$-$$ 4y + 12 = 0 changes to y^{2} = 4ax, then a is equal to

4

BITSAT 2020

MCQ (Single Correct Answer)

+3

-1

The distance of point of intersection of the tangents to the parabola x = 4y $$-$$ y^{2} drawn at the points where it is meet by Y-axis, from its focus is

Questions Asked from Parabola (MCQ (Single Correct Answer))

Number in Brackets after Paper Indicates No. of Questions

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Physics

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Coordinate Geometry

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