1
JEE Advanced 2018 Paper 2 Offline
+3
-1
Let $$H:{{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$, where a > b > 0, be a hyperbola in the XY-plane whose conjugate axis LM subtends an angle of 60$$^\circ$$ at one of its vertices N. Let the area of the $$\Delta$$LMN be $$4\sqrt 3$$.

List - I List - II
P. The length of the conjugate axis of H is 1. 8
Q. The eccentricity of H is 2. $${4 \over {\sqrt 3 }}$$
R. The distance between the foci of H is 3. $${2 \over {\sqrt 3 }}$$
S. The length of the latus rectum of H is 4. 4
A
P $$\to$$ 4 ; Q $$\to$$ 2 ; R $$\to$$ 1 ; S $$\to$$ 3
B
P $$\to$$ 4 ; Q $$\to$$ 3 ; R $$\to$$ 1 ; S $$\to$$ 2
C
P $$\to$$ 4 ; Q $$\to$$ 1 ; R $$\to$$ 3 ; S $$\to$$ 2
D
P $$\to$$ 3 ; Q $$\to$$ 4 ; R $$\to$$ 2 ; S $$\to$$ 1
2
JEE Advanced 2017 Paper 1 Offline
+3
-1
By appropriately matching the information given in the three columns of the following table.

Columns 1, 2 and 3 contain conics, equations of tangents to the conics and points of contact, respectively.

Column - 1 Column - 2 Column - 3
(i) $${x^2} + {y^2} = a$$ $$my = {m^2}x + a$$ $$\left( {{a \over {{m^2}}},\,{{2a} \over m}} \right)$$
(ii) $${x^2}{a^2}{y^2} = {a^2}]$$ $$y = mx + a\sqrt {{m^2} + 1}$$ $$\left( {{{ - ma} \over {\sqrt {{m^2} + 1} }},\,{a \over {\sqrt {{m^2} + 1} }}} \right)$$
(iii) $${y^2} = 4ax$$ $$y = mx + \sqrt {{a^2}{m^2} - 1}$$ $$\left( {{{ - {a^2}m} \over {\sqrt {{a^2}{m^2} + 1} }},\,{1 \over {\sqrt {{a^2}{m^2} + 1} }}} \right)$$
(iv) $${x^2} - {a^2}{y^2} = {a^2}$$ $$y = mx + \sqrt {{a^2}{m^2} + 1}$$ $$\left( {{{ - {a^2}m} \over {\sqrt {{a^2}{m^2} - 1} }},\,{{ - 1} \over {\sqrt {{a^2}{m^2} - 1} }}} \right)$$
For $$a = \sqrt 2$$, if a tangent is drawn to a suitable conic (Column 1) at the point of contact ($$-$$1, 1), then which of the following options is the only CORRECT combination for obtaining its equation?
A
(I) (ii) Q)
B
(I) (ii) (P)
C
(III) (i) (P)
D
(II) (ii) (Q)
3
JEE Advanced 2017 Paper 1 Offline
+3
-1
By appropriately matching the information given in the three columns of the following table.

Columns 1, 2 and 3 contain conics, equations of tangents to the conics and points of contact, respectively.

Column - 1 Column - 2 Column - 3
(i) $${x^2} + {y^2} = a$$ $$my = {m^2}x + a$$ $$\left( {{a \over {{m^2}}},\,{{2a} \over m}} \right)$$
(ii) $${x^2}{a^2}{y^2} = {a^2}]$$ $$y = mx + a\sqrt {{m^2} + 1}$$ $$\left( {{{ - ma} \over {\sqrt {{m^2} + 1} }},\,{a \over {\sqrt {{m^2} + 1} }}} \right)$$
(iii) $${y^2} = 4ax$$ $$y = mx + \sqrt {{a^2}{m^2} - 1}$$ $$\left( {{{ - {a^2}m} \over {\sqrt {{a^2}{m^2} + 1} }},\,{1 \over {\sqrt {{a^2}{m^2} + 1} }}} \right)$$
(iv) $${x^2} - {a^2}{y^2} = {a^2}$$ $$y = mx + \sqrt {{a^2}{m^2} + 1}$$ $$\left( {{{ - {a^2}m} \over {\sqrt {{a^2}{m^2} - 1} }},\,{{ - 1} \over {\sqrt {{a^2}{m^2} - 1} }}} \right)$$
The tangent to a suitable conic (Column 1) at $$\left( {\sqrt 3 ,\,{1 \over 2}} \right)$$ is found to be $$\sqrt 3 x + 2y = 4$$, then which of the following options is the only CORRECT combination?
A
(IV) (iv) (S)
B
(II) (iv) (R)
C
(IV) (iii) (S)
D
(II) (ii) (R)
4
IIT-JEE 2011 Paper 2 Offline
+3
-0.75
Let $$P(6, 3)$$ be a point on the hyperbola $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$. If the normal at the point $$P$$ intersects the $$x$$-axis at $$(9, 0)$$, then the eccentricity of the hyperbola is
A
$$\sqrt {{5 \over 2}}$$
B
$$\sqrt {{3 \over 2}}$$
C
$${\sqrt 2 }$$
D
$${\sqrt 3 }$$
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