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1

### JEE Main 2019 (Online) 12th January Morning Slot

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

## Explanation

f (a) = 2a(12 $$-$$ a)2

f '(a) = 2(12 $$-$$ 3a2)

Maximum at a = 2

maximum area = f(2) = 32
2

### JEE Main 2019 (Online) 11th January Evening Slot

A circle cuts a chord of length 4a on the x-axis and passes through a point on the y-axis, distant 2b from the origin. Then the locus of the centre of this circle, is :
A
an ellipse
B
a parabola
C
a hyperbola
D
a straight line

## Explanation

Let equation of circle is

x2 + y2 + 2fx + 2fy + e = 0, it passes through (0, 2b)

$$\Rightarrow$$  0 + 4b2 + 2g $$\times$$ 0 + 4f + c = 0

$$\Rightarrow$$  4b2 + 4f + c = 0       . . . (i)

$$2\sqrt {{g^2} - c} = 4a$$      . . . (ii)

g2 $$-$$ c = 4a2 $$\Rightarrow$$ c = $$\left( {{g^2} - 4{a^2}} \right)$$

Putting in equation (1)

$$\Rightarrow$$  4b2 + 4f + g2 $$-$$ 4a2 = 0

$$\Rightarrow$$  x2 + 4y + 4(b2 $$-$$ a2) = 0, it represent a hyperbola.
3

### JEE Main 2019 (Online) 11th January Evening Slot

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

## Explanation

Vertex is (a2, 0)

y2 $$=$$ $$-$$(x $$-$$ a2) and x $$=$$ 0 $$\Rightarrow$$ (0, $$\pm$$ 2a)

Area of triangle is $$= {1 \over 2}.$$4a.(a2) = 250

$$\Rightarrow$$  a3 = 125 or a = 5
4

### JEE Main 2019 (Online) 11th January Evening Slot

If a hyperbola has length of its conjugate axis equal to 5 and the distance between its foci is 13, then the eccentricity of the hyperbola is :
A
$${{13} \over 6}$$
B
2
C
$${{13} \over 12}$$
D
$${{13} \over 8}$$

## Explanation

2b = 5 and 2ae = 13

b2 = a2(e2 $$-$$ 1) $$\Rightarrow$$  $${{25} \over 4}$$ = $${{169} \over 4}$$ $$-$$ a2

$$\Rightarrow$$  a $$=$$ 6 $$\Rightarrow$$  e $$=$$ $${{13} \over {12}}$$

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