1
BITSAT 2022
+3
-1

If the straight line $$y = mx + c$$ touches the parabola $${y^2} - 4ax + 4{a^3} = 0$$, then c is

A
$$am + {a \over m}$$
B
$$am - {a \over m}$$
C
$${a \over m} + {a^2}m$$
D
$${a \over m} - {a^2}m$$
2
BITSAT 2022
+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

A
$$2x + y - 12 - 4\sqrt 3 = 0$$
B
$$2x - y - 12 + 4\sqrt 3 = 0$$
C
$$2x - y - 12 - 4\sqrt 3 = 0$$
D
None of these
3
BITSAT 2022
+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)
4
BITSAT 2021
+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
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