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1

### JEE Main 2021 (Online) 27th August Morning Shift

If x2 + 9y2 $$-$$ 4x + 3 = 0, x, y $$\in$$ R, then x and y respectively lie in the intervals :
A
$$\left[ { - {1 \over 3},{1 \over 3}} \right]$$ and $$\left[ { - {1 \over 3},{1 \over 3}} \right]$$
B
$$\left[ { - {1 \over 3},{1 \over 3}} \right]$$ and [1, 3]
C
[1, 3] and [1, 3]
D
[1, 3] and $$\left[ { - {1 \over 3},{1 \over 3}} \right]$$

## Explanation

x2 + 9y2 $$-$$ 4x + 3 = 0

(x2 $$-$$ 4x) + (9y2) + 3 = 0

(x2 $$-$$ 4x + 4) + (9y2) + 3 $$-$$ 4 = 0

(x $$-$$ 2)2 + (3y)2 = 1

$${{{{(x - 2)}^2}} \over {{{(1)}^2}}} + {{{y^2}} \over {{{\left( {{1 \over 3}} \right)}^2}}} = 1$$ (equation of an ellipse).

As it is equation of an ellipse, x & y can vary inside the ellipse.

So, $$x - 2 \in [ - 1,1]$$ and $$y \in \left[ { - {1 \over 3},{1 \over 3}} \right]$$

x $$\in$$ [1, 3] $$y \in \left[ { - {1 \over 3},{1 \over 3}} \right]$$
2

### JEE Main 2021 (Online) 27th August Morning Shift

A tangent and a normal are drawn at the point P(2, $$-$$4) on the parabola y2 = 8x, which meet the directrix of the parabola at the points A and B respectively. If Q(a, b) is a point such that AQBP is a square, then 2a + b is equal to :
A
$$-$$16
B
$$-$$18
C
$$-$$12
D
$$-$$20

## Explanation

Given, parabola

$${y^2} = 8x$$ ...... (i)

Equation of tangent at $$P(2, - 4)$$ is

$$- 4y = 4(x + 2)$$

or, $$x + y + 2 = 0$$ ..... (ii)

and Equation of normal to the parabola is

$$x - y + C = 0$$

$$\therefore$$ Normal passes through $$(2, - 4)$$

$$\therefore$$ $$C = - 6$$

Normal : $$x - y = 6$$ ..... (iii)

Equation of directrix of parabola

$$x = - 2$$ ..... (iv)

Point of intersection of tangent and normal with directrix are $$x = - 2$$ at $$A( - 2,0)$$ and $$B( - 2, - 8)$$ respectively.

$$Q(a,b)$$ and $$P(2, - 4)$$ are given and AQBP is a square.

Mid-point of AB = Mid-point of PQ

$$\Rightarrow ( - 2, - 4) = \left( {{{a + 2} \over 2},{{b - 4} \over 2}} \right) \Rightarrow a = - 6,b = - 4$$

$$\Rightarrow 2a + b = - 16$$

3

### JEE Main 2021 (Online) 26th August Evening Shift

The locus of the mid points of the chords of the hyperbola x2 $$-$$ y2 = 4, which touch the parabola y2 = 8x, is :
A
y3(x $$-$$ 2) = x2
B
x3(x $$-$$ 2) = y2
C
y2(x $$-$$ 2) = x3
D
x2(x $$-$$ 2) = y3

## Explanation

T = S1

xh $$-$$ yk = h2 $$-$$ k2

$$y = {{xh} \over k} - {{({h^2} - {k^2})} \over k}$$

this touches y2 = 8x then $$c = {a \over m}$$

$$\left( {{{{k^2} - {h^2}} \over k}} \right) = {{2k} \over h}$$

2y2 = x(y2 $$-$$ x2)

y2(x $$-$$ 2) = x3
4

### JEE Main 2021 (Online) 26th August Evening Shift

The point $$P\left( { - 2\sqrt 6 ,\sqrt 3 } \right)$$ lies on the hyperbola $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$ having eccentricity $${{\sqrt 5 } \over 2}$$. If the tangent and normal at P to the hyperbola intersect its conjugate axis at the point Q and R respectively, then QR is equal to :
A
$$4\sqrt 3$$
B
6
C
$$6\sqrt 3$$
D
$$3\sqrt 6$$

## Explanation

$$P\left( { - 2\sqrt 6 ,\sqrt 3 } \right)$$ lies on hyperbola

$$\Rightarrow {{24} \over {{a^2}}} - {3 \over {{b^2}}} = 1$$ ...... (i)

$$e = {{\sqrt 5 } \over 2} \Rightarrow {b^2} = {a^2}\left( {{5 \over 4} - 1} \right) \Rightarrow 4{b^2} = {a^2}$$

Put in (i) $$\Rightarrow {6 \over {{b^2}}} - {3 \over {{b^2}}} = 1 \Rightarrow b = \sqrt 3$$

$$\Rightarrow a = \sqrt {12}$$ Tangent at P :

$${{ - x} \over {\sqrt 6 }} - {y \over {\sqrt 3 }} = 1 \Rightarrow Q(0,\sqrt 3 )$$

Slope of $$T = - {1 \over {\sqrt 2 }}$$

Normal at P :

$$y - \sqrt 3 = \sqrt 2 (x + 2\sqrt 6 )$$

$$\Rightarrow R = (0,5\sqrt 3 )$$

$$QR = 6\sqrt 3$$

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