1
BITSAT 2023
MCQ (Single Correct Answer)
+3
-1

If $$y=m_1 x+c_1$$ and $$y=m_2 x+c_2, m_1 \neq m_2$$ are two common tangents of circle $$x^2+y^2=2$$ and parabola $$y^2=x$$, then the value of $$8\left|m_1 m_2\right|$$ is equal to

A
$$7+6 \sqrt{2}$$
B
$$3+4 \sqrt{2}$$
C
$$-5+6 \sqrt{2}$$
D
$$-4+3 \sqrt{2}$$
2
BITSAT 2022
MCQ (Single Correct Answer)
+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
$${a \over m} - {a^2}m$$
3
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

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
4
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)
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