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1

JEE Advanced 2015 Paper 2 Offline

MCQ (More than One Correct Answer)
Consider the hyperbola $$H:{x^2} - {y^2} = 1$$ and a circle $$S$$ with center $$N\left( {{x_2},0} \right)$$. Suppose that $$H$$ and $$S$$ touch each other at a point $$P\left( {{x_1},{y_1}} \right)$$ with $${{x_1} > 1}$$ and $${{y_1} > 0}$$. The common tangent to $$H$$ and $$S$$ at $$P$$ intersects the $$x$$-axis at point $$M$$. If $$(l, m)$$ is the centroid of the triangle $$PMN$$, then the correct expressions(s) is(are)
A
$${{dl} \over {d{x_1}}} = 1 - {1 \over {3x_1^2}}$$ for $${x_1} > 1$$
B
$${{dm} \over {d{x_1}}} = {{{x_1}} \over {3\left( {\sqrt {x_1^2 - 1} } \right)}}$$ for $${x_1} > 1$$
C
$${{dl} \over {d{x_1}}} = 1 + {1 \over {3x_1^2}}$$ for $${x_1} > 1$$
D
$${{dm} \over {d{y_1}}} = {1 \over 3}$$ for $${y_1} > 0$$

Explanation

Equation of family of circles touching hyperbola at (x1, y1) is

(x $$-$$ x1)2 + (y $$-$$ y1)2 + $$\lambda$$(xx1 $$-$$ yy1 $$-$$ 1) =0

Now, its centre is (x2, 0).

$$\therefore$$ $$\left[ {{{ - (\lambda {x_1} - 2{x_1})} \over 2},{{ - ( - 2{y_1} - \lambda {y_1})} \over 2}} \right] = ({x_2},0)$$

$$ \Rightarrow 2{y_1} + \lambda {y_1} = 0 \Rightarrow \lambda = - 2$$

and $$2{x_1} - \lambda x = 2{x_2} \Rightarrow {x_2} = 2{x_1}$$

$$\therefore$$ $$P({x_1},\sqrt {x_1^2 - 1} )$$

and $$N({x_2},0) = (2{x_1},0)$$

As tangent intersect X-axis at $$M\left( {{1 \over x},0} \right)$$.

Centroid of $$\Delta PWN = (l,m)$$

$$ \Rightarrow \left( {{{3{x_1} + {1 \over {{x_1}}}} \over 3},{{{y_1} + 0 + 0} \over 3}} \right) = (l,m)$$

$$ \Rightarrow l = {{3{x_1} + {1 \over {{x_1}}}} \over 3}$$

On differentiating w.r.t. x1, we get

$${{dl} \over {d{x_1}}} = {{3 - {1 \over {x_1^2}}} \over 3}$$

$$ \Rightarrow {{dl} \over {d{x_1}}} = 1 - {1 \over {3x_2^1}}$$, for x1 > 1

and $$m = {{\sqrt {x_1^2 - 1} } \over 3}$$

On differentiating w.r.t. x1, we get

$${{dm} \over {d{x_1}}} = {{2{x_1}} \over {2 \times 3\sqrt {x_1^2 - 1} }} = {{{x_1}} \over {3\sqrt {x_1^2 - 1} }}$$, for x1 > 1

Also, $$m = {{{y_1}} \over 3}$$

On differentiating w.r.t. y1, we get

$${{dm} \over {d{y_1}}} = {1 \over 3}$$, for y1 > 0

2

JEE Advanced 2015 Paper 1 Offline

MCQ (More than One Correct Answer)
Let $$P$$ and $$Q$$ be distinct points on the parabola $${y^2} = 2x$$ such that a circle with $$PQ$$ as diameter passes through the vertex $$O$$ of the parabola. If $$P$$ lies in the first quadrant and the area of the triangle $$\Delta OPQ$$ is $${3\sqrt 2 ,}$$ then which of the following is (are) the coordinates of $$P$$?
A
$$\left( {4,2\sqrt 2 } \right)$$
B
$$\left( {9,3\sqrt 2 } \right)$$
C
$$\left( {{1 \over 4},{1 \over {\sqrt 2 }}} \right)$$
D
$$\left( {1,\sqrt 2 } \right)$$

Explanation

Let $$P\left( {{{t_1^2} \over 2},{t_1}} \right)$$ and $$Q\left( {{{t_2^2} \over 2},{t_2}} \right)$$ be two distinct points on the parabola $${y^2} = 2x$$.

The circle with PQ as diameter passes through the vertex O(0, 0) of the parabola.

Clearly, PO $$\bot$$ OQ

So, slope of PO $$\times$$ slope of OQ = $$-$$1

or, $${{{t_1} - 0} \over {{{t_1^2} \over 2} - 0}} \times {{{t_2} - 0} \over {{{t_2^2} \over 2} - 0}} = - 1$$

or, $${2 \over {{t_1}}} \times {2 \over {{t_2}}} = - 1$$

or, $${t_1}{t_2} = - 4$$

By question, area of $$\Delta OPQ = 3\sqrt 2 $$

or, $${1 \over 2}\left| {\matrix{ 0 & 0 & 1 \cr {{{t_1^2} \over 2}} & {{t_1}} & 1 \cr {{{t_2^2} \over 2}} & {{t_2}} & 1 \cr } } \right| = 3\sqrt 2 $$

or, $${1 \over 2}\left| {{{t_1^2{t_2}} \over 2} - {{{t_1}t_2^2} \over 2}} \right| = 3\sqrt 2 $$

or, $$\left| {{t_1}{t_2}} \right|\left| {{t_1} - {t_2}} \right| = 12\sqrt 2 $$

or, $$\left| {{t_1} + {4 \over {{t_1}}}} \right| = 3\sqrt 2 $$ [$$\because$$ $${t_1}{t_2} = - 4$$]

or, $${t_1} + {4 \over {{t_1}}} = 3\sqrt 2 $$ [$$\because$$ P lies in first quadrant]

or, $$t_1^2 - 3\sqrt 2 {t_1} + 4 = 0$$

or, $${t_1} = {{3\sqrt 2 \pm \sqrt {18 - 4 \times 1 \times 4} } \over 2}$$

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

$$ = 2\sqrt 2 ,\sqrt 2 $$

$$\therefore$$ coordinates of $$P = (4,2\sqrt 2 )$$ or $$(1,\sqrt 2 )$$.

Therefore, (A) and (D) are correct options.

3

JEE Advanced 2015 Paper 2 Offline

MCQ (More than One Correct Answer)
Let $${E_1}$$ and $${E_2}$$ be two ellipses whose centres are at the origin. The major axes of $${E_1}$$ and $${E_2}$$ lie along the $$x$$-axis and the $$y$$-axis, respectively. Let $$S$$ be the circle $${x^2} + {\left( {y - 1} \right)^2} = 2$$. The straight line $$x+y=3$$ touches the curves $$S$$, $${E_1}$$ and $${E_2}$$ at $$P, Q$$ and $$R$$ respectively. Suppose that $$PQ = PR = {{2\sqrt 2 } \over 3}$$. If $${e_1}$$ and $${e_2}$$ are the eccentricities of $${E_1}$$ and $${E_2}$$, respectively, then the correct expression(s) is (are)
A
$$\mathop e\nolimits_1^2 + \mathop e\nolimits_2^2 = {{43} \over {40}}$$
B
$${e_1}{e_2} = {{\sqrt 7 } \over {2\sqrt {10} }}$$
C
$$\left| {\mathop e\nolimits_1^2 + \mathop e\nolimits_2^2 } \right| = {5 \over 8}$$
D
$${e_1}{e_2} = {{\sqrt 3 } \over 4}$$

Explanation

Here, $${E_1}:{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1,\,(a > b)$$

$${E_2}:{{{x^2}} \over {{c^2}}} + {{{y^2}} \over {{d^2}}} = 1,\,(c < d)$$

and $$S:{x^2} + {(y - 1)^2} = 2$$

as tangent to E1, E2 and S is $$x + y = 3$$.

Let the point of contact of tangent be $$({x_1},{y_1})$$ to S.

$$\therefore$$ $$x\,.\,{x_1} + y\,.\,{y_1} - (y + {y_1}) + 1 = 2$$

or $$x{x_1} + y{y_1} - y = (1 + {y_1})$$, same as $$x + y = 3$$.

$$ \Rightarrow {{{x_1}} \over 1} = {{{y_1} - 1} \over 1} = {{1 + {y_1}} \over 3}$$

i.e. $${x_1} = 1$$ and $${y_1} = 2$$

$$\therefore$$ $$P = (1,2)$$

Since, $$PR = PQ = {{2\sqrt 2 } \over 3}$$. Thus, by parametric form,

$${{x - 1} \over { - 1/\sqrt 2 }} = {{y - 2} \over {1/\sqrt 2 }} = \pm {{2\sqrt 2 } \over 3}$$

$$ \Rightarrow \left( {x = {5 \over 3},y = {4 \over 3}} \right)$$

and $$\left( {x = {1 \over 3},y = {8 \over 3}} \right)$$

$$\therefore$$ $$Q = \left( {{5 \over 3},{4 \over 3}} \right)$$ and $$R = \left( {{1 \over 3},{8 \over 3}} \right)$$

Now, equation of tangent at Q on ellipse E1 is

$${{x\,.\,5} \over {{a^2}\,.\,3}} + {{y\,.\,4} \over {{b^2}\,.\,3}} = 1$$

On comparing with x + y = 3, we get

$${a^2} = 5$$ and $${b^2} = 4$$

$$\therefore$$ $$e_1^2 = 1 - {{{b^2}} \over {{a^2}}} = 1 - {4 \over 5} = {1 \over 5}$$ ..... (i)

Also, equation of tangent at R on ellipse E2 is

$${{x\,.\,1} \over {{a^2}\,.\,3}} + {{y\,.\,8} \over {{b^2}\,.\,3}} = 1$$

On comparing with x + y = 3, we get

$${a^2} = 1,\,{b^2} = 8$$

$$\therefore$$ $$e_2^2 = 1 - {{{a^2}} \over {{b^2}}} = 1 - {1 \over 8} = {7 \over 8}$$ ...... (ii)

Now, $$e_1^2\,.\,e_2^2 = {7 \over {40}} \Rightarrow {e_1}{e_2} = {{\sqrt 7 } \over {2\sqrt {10} }}$$

and $$e_1^2 + e_2^2 = {1 \over 5} + {7 \over 8} = {{43} \over {40}}$$

Also, $$\left| {e_1^2 - e_2^2} \right| = \left| {{1 \over 5} - {7 \over 8}} \right| = {{27} \over {40}}$$

4

IIT-JEE 2012 Paper 1 Offline

MCQ (More than One Correct Answer)
Tangents are drawn to the hyperbola $${{{x^2}} \over 9} - {{{y^2}} \over 4} = 1,$$ parallel to the straight line $$2x - y = 1,$$ The points of contact of the tangents on the hyperbola are
A
$$\left( {{9 \over {2\sqrt 2 }},{1 \over {\sqrt 2 }}} \right)$$
B
$$\left( -{{9 \over {2\sqrt 2 }},-{1 \over {\sqrt 2 }}} \right)$$
C
$$\left( {3\sqrt 3 , - 2\sqrt 2 } \right)$$
D
$$\left( -{3\sqrt 3 , 2\sqrt 2 } \right)$$

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