1
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Morning Slot

Axis of a parabola lies along x-axis. If its vertex and focus are at distances 2 and 4 respectively from the origin, on the positive x-axis then which of the following points does not lie on it ?
A
(5, 2$$\sqrt 6$$)
B
(6, 4$$\sqrt 2$$)
C
(8, 6)
D
(4, -4)

Explanation



So the equation of the parabola,

$${\left( {y - 0} \right)^2} = 4.a\left( {x - 2} \right)$$

$$ \Rightarrow $$  y2 = 4.2 (x $$-$$ 2)

$$ \Rightarrow $$  y2 = 8 (x $$-$$ 2)

By checking each options you can see. point (8, 6) does not lie on the parabola.
2
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Morning Slot

Let $$0 < \theta < {\pi \over 2}$$. If the eccentricity of the

hyperbola $${{{x^2}} \over {{{\cos }^2}\theta }} - {{{y^2}} \over {{{\sin }^2}\theta }}$$ = 1 is greater

than 2, then the length of its latus rectum lies in the interval :
A
(3, $$\infty $$)
B
$$\left( {{3 \over 2},2} \right]$$
C
$$\left( {1,{3 \over 2}} \right]$$
D
$$\left( {2,3} \right]$$

Answer

$$\infty $$

Explanation

Given hyperbola,

$${{{x^2}} \over {{{\cos }^2}\theta }} - {{{y^2}} \over {{{\sin }^2}\theta }} = 1$$

here a = cos$$\theta $$

and b = sin$$\theta $$

We know, eccentricity of the hyperbola is,

$$\sqrt {1 + {{{b^2}} \over {{a^2}}}} $$

$$ \therefore $$  Here eccentricity

(e) = $$\sqrt {1 + {{{{\sin }^2}\theta } \over {{{\cos }^2}\theta }}} $$

Given that,

$$\sqrt {1 + {{{{\sin }^2}\theta } \over {{{\cos }^2}\theta }}} > 2$$

$$ \Rightarrow $$  $$\sqrt {1 + {{\tan }^2}\theta } > 2$$

$$ \Rightarrow $$   1 + tan2$$\theta $$ > 4

$$ \Rightarrow $$  tan2$$\theta $$ > 3

$$ \Rightarrow $$  tan$$\theta $$ > $$ \pm \sqrt 3 $$

As given $$\theta $$ $$ \in $$ $$\left( {0,{\pi \over 2}} \right)$$

possible value of tan$$\theta $$ > $$\sqrt 3 $$

So, $$\theta $$ can be in the range $${\pi \over 3} < \theta < {\pi \over 2}$$



We know latus ractum (LR) = $${{2{b^2}} \over a}$$

$$ \therefore $$  LR = $${{2{{\sin }^2}\theta } \over {\cos \theta }}$$

= 2 tan$$\theta $$ sin$$\theta $$

We know in the range $${\pi \over 3} < \theta < {\pi \over 2}$$ tan$$\theta $$ and sin$$\theta $$ both are increasing function.

So, at $${\pi \over 3}$$ value of LR will be minimum and at $${\pi \over 2}$$ value of LR will be maximum.

$$ \therefore $$  Minimum value of LR = 2tan$${\pi \over 3}$$ sin$${\pi \over 3}$$

= 2 $$ \times $$ $$\sqrt 3 \times {{\sqrt 3 } \over 2}$$
= 3

Maximum value of LR = 2tan$${\pi \over 2}$$ sin$${\pi \over 2}$$

= 2$$\left( \infty \right) \times 1$$

= $$\infty $$

$$ \therefore $$  Internal of LR = (3, $$\infty $$)
3
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Morning Slot

Equation of a common tangent to the circle, x2 + y2 – 6x = 0 and the parabola, y2 = 4x is :
A
$$2\sqrt 3 $$y = 12x + 1
B
$$\sqrt 3 $$y = x + 3
C
$$2\sqrt 3 $$y = -x - 12
D
$$\sqrt 3 $$y = 3x + 1

Explanation

We know,

Equation of tangent to the parabola y2 = 4ax is,

y = mx + $${a \over m}$$

$$ \therefore $$  Equation of tangent to the parabola y2 = 4x is,

y = mx + $${1 \over m}$$

$$ \Rightarrow $$  m2x $$-$$ ym + 1 = 0

This tangent is also the tangent to the circle x2 + y2 $$-$$ 6x = 0

So, the perpendicular distance from the center of the circle to the tangent is equal to the radius of the circle.

Here center is at (3, 0) of the circle and radius = 3

$$ \therefore $$  $$\left| {{{3{m^2} + 1} \over {\sqrt {{m^4} + {m^2}} }}} \right| = 3$$

$$ \Rightarrow $$  (3m2 + 1)2 = 9(m4 + m2)

$$ \Rightarrow $$  9m4 + 6m2 + 1 = 9m4 + 9m2

$$ \Rightarrow $$  3m2 = 1

$$ \Rightarrow $$  m = $$ \pm $$ $${1 \over {\sqrt 3 }}$$

So, possible tangents are

y = $${1 \over {\sqrt 3 }}$$x + $$\sqrt 3 $$

$$ \Rightarrow $$  $$\sqrt 3 $$y = x + 3

or   y = $$-$$ $${x \over {\sqrt 3 }}$$ $$-$$ $$\sqrt 3 $$

$$ \Rightarrow $$  $$\sqrt 3 y$$ = $$-$$ x $$-$$ 3
4
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Morning Slot

If $$\theta $$ denotes the acute angle between the curves, y = 10 – x2 and y = 2 + x2 at a point of their intersection, the |tan $$\theta $$| is equal to :
A
$$8 \over 15$$
B
$$4 \over 9$$
C
$$7 \over 17$$
D
$$8 \over 17$$

Explanation



Angle between the curves is the acute angle between the tangents at the point of intersection.

y = 10 $$-$$ x2 (for curve 1)

and y = 2 + x2 (for curve 2)

$$ \therefore $$  10 $$-$$ x2 = 2 + x2

$$ \Rightarrow $$  2x2 = 8

$$ \Rightarrow $$  x2 = 4

$$ \Rightarrow $$  x = 2, $$-$$ 2

$$ \therefore $$  points of intersection (2, 6) and ($$-$$ 2, 6)

$${{dy} \over {dx}}$$ for curve 1 = $$-$$ 2x

$$ \therefore $$  Slope(m1) of curve 1 is = $$-$$ 2(2) = $$-$$ 4

$${{dy} \over {dx}}$$ for curve 2 = 2x

$$ \therefore $$  slope (m2) of curve 2 = 2 $$ \times $$ 2 = 4

$$ \therefore $$  tan$$\theta $$ = $$\left| {{{{m_1} - {m_2}} \over {1 + {m_1}{m_2}}}} \right|$$

= $$\left| {{{ - 4 - 4} \over {1 + \left( { - 16} \right)}}} \right|$$

= $${8 \over {15}}$$

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