1
MCQ (Single Correct Answer)

JEE Main 2018 (Online) 16th April Morning Slot

The locus of the point of intersection of the lines, $$\sqrt 2 x - y + 4\sqrt 2 k = 0$$ and $$\sqrt 2 k\,x + k\,y - 4\sqrt 2 = 0$$ (k is any non-zero real parameter), is :
A
an ellipse whose eccentricity is $${1 \over {\sqrt 3 }}.$$
B
an ellipse with length of its major axis $$8\sqrt 2 .$$
C
a hyperbola whose eccentricity is $$\sqrt 3 .$$
D
a hyperbola with length of its transverse axis $$8\sqrt 2 .$$

Explanation

Here, lines are :

$$\sqrt 2 x$$ $$-$$ y + 4$$\sqrt 2 k$$ = 0

$$ \Rightarrow $$$$\,\,\,$$ $$\sqrt 2 x + 4\sqrt 2 k = y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....$$(i)

and   $$\sqrt 2 kx + ky - 4\sqrt 2 = 0\,\,\,\,\,...\left( {ii} \right)$$

Put the value of y from (i) in (ii) we get;

$$ \Rightarrow $$2$$\sqrt 2 $$kx + 4$$\sqrt 2 $$(k2 $$-$$ 1) = 0

$$ \Rightarrow $$ x = $${{2\left( {1 - {k^2}} \right)} \over k}$$, y = $${{2\sqrt 2 \left( {1 + {k^2}} \right)} \over k}$$

$$\therefore\,\,\,$$ $${\left( {{y \over {4\sqrt 2 }}} \right)^2} - {\left( {{x \over 4}} \right)^2} = 1$$

$$\therefore\,\,\,$$ length of transverse axis

2a = 2 $$ \times $$ 4$${\sqrt 2 }$$ = 8$${\sqrt 2 }$$

Hence, the locus is a hyperbola with length of its transverse axis equal to 8$${\sqrt 2 }$$
2
MCQ (Single Correct Answer)

JEE Main 2018 (Online) 16th April Morning Slot

If the length of the latus rectum of an ellipse is 4 units and the distance between a focus an its nearest vertex on the major axis is $${3 \over 2}$$ units, then its eccentricity is :
A
$${1 \over 2}$$
B
$${1 \over 3}$$
C
$${2 \over 3}$$
D
$${1 \over 9}$$

Explanation

If the cordinate of focus and vertex are (ae, 0) and (a, 0) respectively,

then distance between focus and vertex,

a $$-$$ ae = $${3 \over 2}$$ (given)

$$ \Rightarrow $$ $$\,\,\,$$ a (1 $$-$$ e) = $${3 \over 2}$$

Length of latus rectum,

$${{2{b^2}} \over a} = 4$$

$$ \Rightarrow $$ $$\,\,\,$$ b2 = 2a

$$ \Rightarrow $$ $$\,\,\,$$ a2(1 $$-$$ e2) = 2a   [As b2 = a2 (1 $$-$$ e2)]

$$ \Rightarrow $$ $$\,\,\,$$ a (1 $$-$$ e) ( 1 + e) = 2

Putting    a (1 $$-$$ e) = $${3 \over 2}$$

$$ \Rightarrow $$ $$\,\,\,$$ $${3 \over 2}$$ (1 + e) = 2

$$ \Rightarrow $$ $$\,\,\,$$ 3 + 3e = 4

$$ \Rightarrow $$ $$\,\,\,$$ e = $${1 \over 3}$$
3
MCQ (Single Correct Answer)

JEE Main 2018 (Online) 16th April Morning Slot

Let P be a point on the parabola, x2 = 4y. If the distance of P from the center of the circle, x2 + y2 + 6x + 8 = 0 is minimum, then the equation of the tangent to the parabola at P, is :
A
x + 4y $$-$$ 2 = 0
B
x $$-$$ y + 3 = 0
C
x + y +1 = 0
D
x + 2y = 0

Explanation

Let P(2t, t2) be any point on the parabola.

Center of the given circle C = ($$-$$ g, $$-$$f) = ($$-$$3, 0)

For PC to be minimum, it must be the normal to the parabola at P.

Slope of line PC = $${{{y_2} - {y_1}} \over {{x_2} - {x_1}}}$$ = $${{{t^2} - 0} \over {2t + 3}}$$

Also, slope of tangent to parabola at P = $${{dy} \over {dx}}$$ = $${x \over 2}$$ = t

$$ \therefore $$ Slope of normal = $${{ - 1} \over t}$$

$$ \therefore $$ $${{{t^2} - 0} \over {2t + 3}}$$ = $${{ - 1} \over t}$$

$$ \Rightarrow $$ t3 + 2t + 3 = 0

$$ \Rightarrow $$ (t+1) (t2 $$-$$ t + 3) = 0

$$\therefore\,\,\,$$ Real roots of above equation is

t = $$-$$ 1

Coordinate of P = (2t, t2) = ($$-$$2, 1)

Slope of tangent to parabola at P = t = $$-$$ 1

Therefore, equation of tangent is :

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

$$ \Rightarrow $$ x + y + 1 = 0
4
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.

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