1
JEE Main 2019 (Online) 12th January Morning Slot
+4
-1
The maximum area (in sq. units) of a rectangle having its base on the x-axis and its other two vertices on the parabola, y = 12 – x2 such that the rectangle lies inside the parabola, is :
A
36
B
20$$\sqrt 2$$
C
18$$\sqrt 3$$
D
32
2
JEE Main 2019 (Online) 12th January Morning Slot
+4
-1
Let P(4, –4) and Q(9, 6) be two points on the parabola, y2 = 4x and let x be any point on the arc POQ of this parabola, where O is the vertex of this parabola, such that the area of $$\Delta$$PXQ is maximum. Then this maximum area (in sq. units) is :
A
$${{625} \over 4}$$
B
$${{125} \over 4}$$
C
$${{75} \over 2}$$
D
$${{125} \over 2}$$
3
JEE Main 2019 (Online) 11th January Evening Slot
+4
-1
If the area of the triangle whose one vertex is at the vertex of the parabola, y2 + 4(x – a2) = 0 and the othertwo vertices are the points of intersection of the parabola and y-axis, is 250 sq. units, then a value of 'a' is :
A
$$5\sqrt 5$$
B
$${\left( {10} \right)^{2/3}}$$
C
$$5\left( {{2^{1/3}}} \right)$$
D
5
4
JEE Main 2019 (Online) 10th January Evening Slot
+4
-1
The length of the chord of the parabola x2 $$=$$ 4y having equation x – $$\sqrt 2 y + 4\sqrt 2 = 0$$  is -
A
$$8\sqrt 2$$
B
$$6\sqrt 3$$
C
$$3\sqrt 2$$
D
$$2\sqrt {11}$$
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