An equilateral triangle OAB is inscribed in the parabola y2=4x with the vertex O at the vertex of the parabola. Then the minimum distance of the circle having AB as a diameter from the origin is
Held on 23 Jan 2026 · Verified 6 Jul 2026.
2(8−33)
2(3+3)
4(6+3)
4(3−3)
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Let the image of parabola $x^{2}=4 y$, in the line $x-y=1$ be $(y+a)^{2}=b(x-c)$, $a, b, c \in \mathrm{~N}$. Then $a+b+c$ is equal to
The distance between the points (3, 4) and (6, 8) is:
If the chord joining the points $\mathrm{P}_{1}\left(x_{1}, y_{1}\right)$ and $\mathrm{P}_{2}\left(x_{2}, y_{2}\right)$ on the parabola $y^{2}=12 x$ subtends a right angle at the vertex of the parabola, then $x_{1} x_{2}-y_{1} y_{2}$ is equal to
The distance between the parallel lines 3x + 4y - 7 = 0 and 3x + 4y + 8 = 0 is:
Let a point $A$ lie between the parallel lines $L_{1}$ and $L_{2}$ such that its distances from $L_{1}$ and $L_{2}$ are 6 and 3 units, respectively. Then the area (in sq. units) of the equilateral triangle $A B C$, where the points $B$ and C lie on the lines $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$, respectively, is :
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