1
BITSAT 2022
MCQ (Single Correct Answer)
+3
-1

For each parabola y = x2 + px + q, meeting coordinate axes at 3-distinct points, if circles are drawn through these points, then the family of circles must pass through

A
(1, 0)
B
(0, 1)
C
(1, 1)
D
(p, q)
2
BITSAT 2021
MCQ (Single Correct Answer)
+3
-1

The origin is shifted to (1, 2). The equation y2 $$-$$ 8x $$-$$ 4y + 12 = 0 changes to y2 = 4ax, then a is equal to

A
1
B
2
C
$$-$$2
D
$$-$$1
3
BITSAT 2020
MCQ (Single Correct Answer)
+3
-1

The distance of point of intersection of the tangents to the parabola x = 4y $$-$$ y2 drawn at the points where it is meet by Y-axis, from its focus is

A
$${{11} \over 4}$$
B
$${{17} \over 4}$$
C
$${{13} \over 4}$$
D
3
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