1
MCQ (Single Correct Answer)

JEE Main 2017 (Online) 9th April Morning Slot

The eccentricity of an ellipse having centre at the origin, axes along the co-ordinate axes and passing through the points (4, −1) and (−2, 2) is :
A
$${1 \over 2}$$
B
$${2 \over {\sqrt 5 }}$$
C
$${{\sqrt 3 } \over 2}$$
D
$${{\sqrt 3 } \over 4}$$

Explanation

Centre at (0, 0)

$${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}}$$ = 1

at point (4, $$-$$ 1)

$${{16} \over {{a^2}}} + {1 \over {{b^2}}}$$ = 1

$$ \Rightarrow $$   16b2 + a2 = a2b2         . . . .(i)

at point ($$-$$ 2, 2)

$${4 \over {{a^2}}} + {4 \over {{b^2}}} = 1$$

$$ \Rightarrow $$   4b2 + 4a2 = a2b2          . . . .(ii)

$$ \Rightarrow $$   16b2 + a2 = 4a2 + 4b2

From equations (i) and (ii)

$$ \Rightarrow $$   3a2 = 12b2

$$ \Rightarrow $$   a2 = 4b2

b2 = a2(1 $$-$$ e2)

$$ \Rightarrow $$   e2 = $${3 \over 4}$$

$$ \Rightarrow $$   e = $${{\sqrt 3 } \over 2}$$
2
MCQ (Single Correct Answer)

JEE Main 2018 (Offline)

Tangents are drawn to the hyperbola 4x2 - y2 = 36 at the points P and Q.

If these tangents intersect at the point T(0, 3) then the area (in sq. units) of $$\Delta $$PTQ is :
A
$$36\sqrt 5 $$
B
$$45\sqrt 5 $$
C
$$54\sqrt 3 $$
D
$$60\sqrt 3 $$

Explanation



Here PQ is the chord of contact.

Equation of PQ is

x.0 $$-$$ y.3 = 36

$$ \Rightarrow \,\,\,\,y = - 12$$

Putting value of y = $$-$$ 12 in the equation

4x2 $$-$$ y2 = 36

we get , 4x2 $$-$$ 144 = 36

$$ \Rightarrow \,\,\,\,4{x^2}\, = \,180$$

$$ \Rightarrow \,\,\,\,\,{x^2}\,\, = \,\,45$$

$$ \Rightarrow \,\,\,\,x = \pm \,3\sqrt 5 $$

$$\therefore\,\,\,$$ Coordinate of $$P = \left( { - 3\sqrt 5 , - 12} \right)$$ and

Coordinate of $$Q = \left( {3\sqrt 5 , - 12} \right)$$

$$\therefore\,\,\,$$ Length of PQ $$ = 3\sqrt 5 + 3\sqrt 5 $$

$$ = 6\sqrt 5 .$$

TM is the height of the triangle Length of TM = 12 + 3 = 15

$$\therefore\,\,\,$$ Area of $$\Delta PQT$$ = $${1 \over 2} \times 6\sqrt 5 \times 15$$

$$ = 45\sqrt 5 $$ sq. units
3
MCQ (Single Correct Answer)

JEE Main 2018 (Offline)

Tangent and normal are drawn at P(16, 16) on the parabola y2 = 16x, which intersect the axis of the parabola at A and B, respectively. If C is the centre of the circle through the points P, A and B and $$\angle $$CPB = $$\theta $$, then a value of tan$$\theta $$ is :
A
$${4 \over 3}$$
B
$${1 \over 2}$$
C
2
D
3

Explanation



As equation of tangent PA at (x1, y1) on the parabola y2 = 4ax,

yy1 = 2a (x + x1)

here (x1, y1) = ( 16, 16)

y . 16 = 2.4 (x + 16)

$$ \Rightarrow $$ 2y = x + 16 .....(1)

At pont A value of y = 0

putting y = 0 in equation (1) we get,

0 = x + 16

$$ \Rightarrow $$ x = $$-$$ 16

$$\therefore\,\,\,$$ Coordinate of point A = ($$-$$ 16, 0)

Slope of line P A :

As$$\,\,\,\,$$ 2y = x + 16

$$\therefore\,\,\,$$ y = $${1 \over 2}\,$$ x + 8

$$\therefore\,\,\,$$ Slope (m) = $${1 \over 2}\,$$

Let slope of perpendicular line PB passing through point p(16, 16) = m'

$$\therefore\,\,\,$$ m m' = $$-$$ 1

$$ \Rightarrow $$ $${1 \over 2}$$ $$ \times $$ m' = $$-$$ 1

$$ \Rightarrow $$ ' = $$-$$ 2

As Equation of normal PB, when slope is m,

y = mx $$-$$ 2am $$-$$am3

Here m = m' = $$-$$ 2 and a = 4

$$\therefore\,\,\,$$ y = $$-$$2x $$-$$ 2(4) ($$-$$2) $$-$$ 4 . ($$-$$2)3

$$ \Rightarrow $$ y = $$-$$ 2x + 16 + 32

$$ \Rightarrow $$ y = $$-$$ 2x + 48 ..... (2)

At point B, y = 0

puttig y = 0 at equation (2) we get,

0 = $$-$$ 2x + 48

$$ \Rightarrow $$x = 24

$$\therefore\,\,\,$$ Coordinate of point B = (24, 0)



A circle is passing through point P, A and B, and C is the center of the circle.

So, AC and BC are the radius.

Then AC = BC, So C is the middle point of line AB.

$$\therefore\,\,\,$$ C = $$\left( {{{24 - 16} \over 2},{{0 + 0} \over 2}} \right)$$ = (4, 0)

$$\angle $$CPB = $$\theta $$ and we have to find tan $$\theta $$.

Slope of line PC = $${{16 - 0} \over {16 - 4}}$$ = $${4 \over 3}$$ =m1

and we know slope of line PB = $$-$$2 = m2

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

$$ \Rightarrow $$ tan $$\theta $$ = $$\left| {{{{4 \over 3} + 2} \over {1 - \left( {{4 \over 3}} \right)\left( 2 \right)}}} \right|$$

$$ \Rightarrow $$ tan $$\theta $$ = $$\left| {{{{{10} \over 3}} \over { - {5 \over 3}}}} \right|$$

$$ \Rightarrow $$ tan $$\theta $$ = $$\left| { - 2} \right|$$

$$ \Rightarrow $$ tan $$\theta $$ = 2
4
MCQ (Single Correct Answer)

JEE Main 2018 (Online) 15th April Morning Slot

If the tangents drawn to the hyperbola 4y2 = x2 + 1 intersect the co-ordinate axes at the distinct points A and B then the locus of the mid point of AB is :
A
x2 $$-$$ 4y2 + 16x2y2 = 0
B
x2 $$-$$ 4y2 $$-$$ 16x2y2 = 0
C
4x2 $$-$$ y2 + 16x2y2 = 0
D
4x2 $$-$$ y2 $$-$$ 16x2y2 = 0

Explanation

Equation of hyperbola is :

4y2 = x2 + 1 $$ \Rightarrow $$ $$-$$ x2 + 4y2 = 1

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

$$ \therefore $$  a = 1, b = $${1 \over 2}$$

Now, tangent to the curve at point (x1, y1) is given by

4 $$ \times $$ 2y1$${{dy} \over {dx}}$$ = 2x1

$$ \Rightarrow $$   $${{dy} \over {dx}}$$ = $${{2{x_1}} \over {8{y_1}}}$$ = $${{{x_1}} \over {4{y_1}}}$$

Equation of tangent at (x1, y1) is

y = mx + c

$$ \Rightarrow $$   y = $${{{x_1}} \over {4{y_1}}}$$ . x + c

As tangent passes through (x1, y1)

$$ \therefore $$   y1 = $${{{x_1}{x_1}} \over {4{y_1}}} + c$$

$$ \Rightarrow $$   C = $${{4y_1^2 - x_1^2} \over {4{y_1}}}$$ = $${1 \over {4{y_1}}}$$

Therefore, y = $${{{x_1}} \over {4{y_1}}}x + {1 \over {4{y_1}}}$$

$$ \Rightarrow $$ 4y1y = x1x + 1

which intersects x axis at A $$\left( {{{ - 1} \over {{x_1}}},0} \right)$$ and y axis at $$B\left( {0,{1 \over {4{y_1}}}} \right)$$

Let midpoint of AB is (h, k)

$$ \therefore $$   h = $${{ - 1} \over {2{x_1}}}$$

$$ \Rightarrow $$$$\,\,\,$$ x1 = $${{ - 1} \over {2h}}$$ & y1 = $${1 \over {8k}}$$

Thus, 4$${\left( {{1 \over {8k}}} \right)^2}$$ = $${\left( {{{ - 1} \over {2h}}} \right)^2}$$ + 1

$$ \Rightarrow $$$$\,\,\,$$ $${1 \over {16{k^2}}}$$ = $${1 \over {4{h^2}}}$$ + 1

$$ \Rightarrow $$ $$\,\,\,$$ 1 = $${{16{k^2}} \over {4{h^2}}}$$ + 16k2

$$ \Rightarrow $$$$\,\,\,$$ h2 = 4k2 + 16h2 k.

So, required equation is

x2 $$-$$ 4y2 $$-$$ 16x2 y2 = 0

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