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JEE Mathematics · 90 questions · Page 2 of 9 · Click an option or "Show Solution" to reveal answer

Q11
A ray of light through (2, 1) is reflected at a point P on the y-axis and then passes through the point (5, 3). If this reflected ray is the directrix of an ellipse with eccentricity 13{1 \over 3} and the distance of the nearer focus from this directrix is 853{8 \over {\sqrt {53} }}, then the equation of the other directrix can be :
A 11x + 7y + 8 = 0 or 11x + 7y −- 15 = 0
B 11x −- 7y −- 8 = 0 or 11x + 7y + 15 = 0
C 2x −- 7y + 29 = 0 or 2x −- 7y −- 7 = 0
D 2x −- 7y −- 39 = 0 or 2x −- 7y −- 7 = 0
Correct Answer
Option C
Solution

Equation of reflected Ray

y−1=27(x+2)y - 1 = {2 \over 7}(x + 2)
7x−7=2x+47x - 7 = 2x + 4
2x−7y+11=02x - 7y + 11 = 0

Let the equation of other directrix is

2x−7y+λ2x - 7y + \lambda

= 0 Distance of directrix from focus

ae−e=853{a \over e} - e = {8 \over {\sqrt {53} }}
3a−a3=8533a - {a \over 3} = {8 \over {\sqrt {53} }}

or

a=353a = {3 \over {\sqrt {53} }}

Distance from other focus

ae+ae{a \over e} + ae
3a+a3=10a3=103×353=10533a + {a \over 3} = {{10a} \over 3} = {{10} \over 3} \times {3 \over {\sqrt {53} }} = {{10} \over {\sqrt {53} }}

Distance between two directrix =

=2ae= {{2a} \over e}
=2×3×353=1853= 2 \times 3 \times {3 \over {\sqrt {53} }} = {{18} \over {\sqrt {53} }}
∣λ−1153∣=1853\left| {{{\lambda - 11} \over {\sqrt {53} }}} \right| = {{18} \over {\sqrt {53} }}
λ−11=18\lambda - 11 = 18

or −- 18

λ=29\lambda = 29

or −-7

2x−7y−7=02x - 7y - 7 = 0

or

2x−7y+29=02x - 7y + 29 = 0
Q12
On the ellipse x28+y24=1{{{x^2}} \over 8} + {{{y^2}} \over 4} = 1 let P be a point in the second quadrant such that the tangent at P to the ellipse is perpendicular to the line x + 2y = 0. Let S and S' be the foci of the ellipse and e be its eccentricity. If A is the area of the triangle SPS' then, the value of (5 −- e2). A is :
A 6
B 12
C 14
D 24
Correct Answer
Option A
Solution

Equation of tangent : y = 2x + 6 at P ∴\therefore P(−-8/3, 2/3)

e=12e = {1 \over {\sqrt 2 }}

S & S' = (−-2, 0) & (2, 0) Area of

Δ\Delta

SPS' =

12×4×23{1 \over 2} \times 4 \times {2 \over 3}

A =

43{4 \over 3}

∴\therefore (5 −- e2)A =

(5−12)\left( {5 - {1 \over 2}} \right)
43{4 \over 3}

= 6

Q13
If x2 + 9y2 −- 4x + 3 = 0, x, y ∈\in R, then x and y respectively lie in the intervals :
A [−13,13]\left[ { - {1 \over 3},{1 \over 3}} \right] and [−13,13]\left[ { - {1 \over 3},{1 \over 3}} \right]
B [−13,13]\left[ { - {1 \over 3},{1 \over 3}} \right] and [1, 3]
C [1, 3] and [1, 3]
D [1, 3] and [−13,13]\left[ { - {1 \over 3},{1 \over 3}} \right]
Correct Answer
Option D
Solution

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(1)2+y2(13)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∈[−1,1]x - 2 \in [ - 1,1]

and

y∈[−13,13]y \in \left[ { - {1 \over 3},{1 \over 3}} \right]

x

∈\in

[1, 3]

y∈[−13,13]y \in \left[ { - {1 \over 3},{1 \over 3}} \right]
Q14
The line 12xcos⁡θ+5ysin⁡θ=6012x\cos \theta + 5y\sin \theta = 60 is tangent to which of the following curves?
A x2 + y2 = 169
B 144x2 + 25y2 = 3600
C 25x2 + 12y2 = 3600
D x2 + y2 = 60
Correct Answer
Option B
Solution
12xcos⁡θ+5ysin⁡θ=6012x\cos \theta + 5y\sin \theta = 60
xcos⁡θ5+ysin⁡θ12=1{{x\cos \theta } \over 5} + {{y\sin \theta } \over {12}} = 1

is tangent to

x225+y2144=1{{{x^2}} \over {25}} + {{{y^2}} \over {144}} = 1
144x2+25y2=3600144{x^2} + 25{y^2} = 3600
Q15
The locus of mid-points of the line segments joining (−-3, −-5) and the points on the ellipse x24+y29=1{{{x^2}} \over 4} + {{{y^2}} \over 9} = 1 is :
A 9x2+4y2+18x+8y+145=09{x^2} + 4{y^2} + 18x + 8y + 145 = 0
B 36x2+16y2+90x+56y+145=036{x^2} + 16{y^2} + 90x + 56y + 145 = 0
C 36x2+16y2+108x+80y+145=036{x^2} + 16{y^2} + 108x + 80y + 145 = 0
D 36x2+16y2+72x+32y+145=036{x^2} + 16{y^2} + 72x + 32y + 145 = 0
Correct Answer
Option C
Solution

General point on

x24+y29=1{{{x^2}} \over 4} + {{{y^2}} \over 9} = 1

is A(2cosθ\theta, 3sinθ\theta) given B(−-3, −-5) midpoint

C(2cos⁡θ−32,3sin⁡θ−52)C\left( {{{2\cos \theta - 3} \over 2},{{3\sin \theta - 5} \over 2}} \right)
h=2cos⁡θ−32;k=3sin⁡θ−52h = {{2\cos \theta - 3} \over 2};k = {{3\sin \theta - 5} \over 2}
⇒(2h+32)2+(2k+53)2=1\Rightarrow {\left( {{{2h + 3} \over 2}} \right)^2} + {\left( {{{2k + 5} \over 3}} \right)^2} = 1
⇒36x2+16y2+108x+80y+145=0\Rightarrow 36{x^2} + 16{y^2} + 108x + 80y + 145 = 0
Q16
An angle of intersection of the curves, x2a2+y2b2=1{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1 and x2 + y2 = ab, a > b, is :
A tan⁡−1(a+bab){\tan ^{ - 1}}\left( {{{a + b} \over {\sqrt {ab} }}} \right)
B tan⁡−1(a−b2ab){\tan ^{ - 1}}\left( {{{a - b} \over {2\sqrt {ab} }}} \right)
C tan⁡−1(a−bab){\tan ^{ - 1}}\left( {{{a - b} \over {\sqrt {ab} }}} \right)
D tan⁡−1(2ab){\tan ^{ - 1}}\left( {2\sqrt {ab} } \right)
Correct Answer
Option C
Solution
x2a2+y2b2=1,x2+y2=ab{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1,{x^2} + {y^2} = ab
2x1a2+2y1y′b2=0{{2{x_1}} \over {{a^2}}} + {{2{y_1}y'} \over {{b^2}}} = 0
⇒y1′=−x1a2b2y1\Rightarrow {y_1}' = {{ - {x_1}} \over {{a^2}}}{{{b^2}} \over {{y_1}}}

.... (1) ∴\therefore

2x1+2y1y′=02{x_1} + 2{y_1}y' = 0
⇒y2′=−x1y1\Rightarrow {y_2}' = {{ - {x_1}} \over {{y_1}}}

..... (2) Here (x1y1) is point of intersection of both curves ∴\therefore

x12=a2ba+b,y12=ab2a+bx_1^2 = {{{a^2}b} \over {a + b}},y_1^2 = {{a{b^2}} \over {a + b}}

∴\therefore

tan⁡θ=∣y1′−y2′1+y1′y2′∣=∣−x1b2a2y1+x1y11+x12b2a2y12∣\tan \theta = \left| {{{{y_1}' - {y_2}'} \over {1 + {y_1}'{y_2}'}}} \right| = \left| {{{{{ - {x_1}{b^2}} \over {{a^2}{y_1}}} + {{{x_1}} \over {{y_1}}}} \over {1 + {{x_1^2{b^2}} \over {{a^2}y_1^2}}}}} \right|
tan⁡θ=∣−b2x1y1+a2x1y1a2y12+b2x12∣\tan \theta = \left| {{{ - {b^2}{x_1}{y_1} + {a^2}{x_1}{y_1}} \over {{a^2}y_1^2 + {b^2}x_1^2}}} \right|
tan⁡θ=∣a−bab∣\tan \theta = \left| {{{a - b} \over {\sqrt {ab} }}} \right|
Q17
Let θ\theta be the acute angle between the tangents to the ellipse x29+y21=1{{{x^2}} \over 9} + {{{y^2}} \over 1} = 1 and the circle x2+y2=3{x^2} + {y^2} = 3 at their point of intersection in the first quadrant. Then tanθ\theta is equal to :
A 523{5 \over {2\sqrt 3 }}
B 23{2 \over {\sqrt 3 }}
C 43{4 \over {\sqrt 3 }}
D 2
Correct Answer
Option B
Solution

The point of intersection of the curves

x29+y21=1{{{x^2}} \over 9} + {{{y^2}} \over 1} = 1

and

x2+y2=3{x^2} + {y^2} = 3

in the first quadrant is

(32,32)\left( {{3 \over 2},{{\sqrt 3 } \over 2}} \right)

Now slope of tangent to the ellipse

x29+y21=1{{{x^2}} \over 9} + {{{y^2}} \over 1} = 1

at

(32,32)\left( {{3 \over 2},{{\sqrt 3 } \over 2}} \right)

is

m1=−133{m_1} = - {1 \over {3\sqrt 3 }}

And slope of tangent to the circle at

(32,32)\left( {{3 \over 2},{{\sqrt 3 } \over 2}} \right)

is m2

=−3= - \sqrt 3

So, if angle between both curves is θ\theta then

tan⁡θ=∣m1−m21+m1m2∣=∣−133+31+(−133(−3))∣\tan \theta = \left| {{{{m_1} - {m_2}} \over {1 + {m_1}{m_2}}}} \right| = \left| {{{ - {1 \over {3\sqrt 3 }} + \sqrt 3 } \over {1 + \left( { - {1 \over {3\sqrt 3 }}\left( { - \sqrt 3 } \right)} \right)}}} \right|
=23= {2 \over {\sqrt 3 }}

Option (b)

Q18
Let the eccentricity of an ellipse x2a2+y2b2=1{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1, a>ba > b, be 14{1 \over 4}. If this ellipse passes through the point (−425,3)\left( { - 4\sqrt {{2 \over 5}} ,3} \right), then a2+b2{a^2} + {b^2} is equal to :
A 29
B 31
C 32
D 34
Correct Answer
Option B
Solution
x2a2+y2b2=1{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1
⇒(−425)2a2+32b2=1\Rightarrow {{{{\left( { - 4\sqrt {{2 \over 5}} } \right)}^2}} \over {{a^2}}} + {{32} \over {{b^2}}} = 1
⇒325a2+9b2=1\Rightarrow {{32} \over {5{a^2}}} + {9 \over {{b^2}}} = 1

..... (i)

a2(1−e2)=b2{a^2}(1 - {e^2}) = {b^2}
a2(1−116)=b2{a^2}\left( {1 - {1 \over {16}}} \right) = {b^2}
15a2=16b2⇒a2=16b21515{a^2} = 16{b^2} \Rightarrow {a^2} = {{16{b^2}} \over {15}}

From (i)

6b2+9b2=1⇒b2=15{6 \over {{b^2}}} + {9 \over {{b^2}}} = 1 \Rightarrow {b^2} = 15

&

a2=16{a^2} = 16
a2+b2=15+16=31{a^2} + {b^2} = 15 + 16 = 31
Q19
If m is the slope of a common tangent to the curves x216+y29=1{{{x^2}} \over {16}} + {{{y^2}} \over 9} = 1 and x2+y2=12{x^2} + {y^2} = 12, then 12m212{m^2} is equal to :
A 6
B 9
C 10
D 12
Correct Answer
Option B
Solution
C1:x216+y29=1{C_1}:{{{x^2}} \over {16}} + {{{y^2}} \over 9} = 1

and

C2:x2+y2=12{C_2}:{x^2} + {y^2} = 12

Let

y=mx± 16m2+9y = mx \pm \,\sqrt {16{m^2} + 9}

be any tangent to C1 and if this is also tangent to C2 then

∣16m2+9m2+1∣=12\left| {{{\sqrt {16{m^2} + 9} } \over {\sqrt {{m^2} + 1} }}} \right| = \sqrt {12}
⇒16m2+9=12m2+12\Rightarrow 16{m^2} + 9 = 12{m^2} + 12
⇒4m2=3⇒12m2=9\Rightarrow 4{m^2} = 3 \Rightarrow 12{m^2} = 9
Q20
The locus of the mid point of the line segment joining the point (4, 3) and the points on the ellipse x2+2y2=4{x^2} + 2{y^2} = 4 is an ellipse with eccentricity :
A 32{{\sqrt 3 } \over 2}
B 122{1 \over {2\sqrt 2 }}
C 12{1 \over {\sqrt 2 }}
D 12{1 \over 2}
Correct Answer
Option C
Solution

Let

P(2cos⁡θ, 2sin⁡θ)P(2\cos \theta ,\,\sqrt 2 \sin \theta )

be any point on ellipse

x24+y22=1{{{x^2}} \over 4} + {{{y^2}} \over 2} = 1

and Q(4, 3) and let (h, k) be the mid point of PQ then

h=2cos⁡θ+42, k=2sin⁡θ+32h = {{2\cos \theta + 4} \over 2},\,k = {{\sqrt 2 \sin \theta + 3} \over 2}

∴\therefore

cos⁡θ=h−2, sin⁡θ=2k−32\cos \theta = h - 2,\,\sin \theta = {{2k - 3} \over {\sqrt 2 }}

∴\therefore

(h−2)2+(2k−32)2=1{(h - 2)^2} + {\left( {{{2k - 3} \over {\sqrt 2 }}} \right)^2} = 1
⇒(x−2)21+(y−32)212=1\Rightarrow {{{{(x - 2)}^2}} \over 1} + {{{{\left( {y - {3 \over 2}} \right)}^2}} \over {{1 \over 2}}} = 1

∴\therefore

e=1−12=12e = \sqrt {1 - {1 \over 2}} = {1 \over {\sqrt 2 }}
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