1
JEE Advanced 2018 Paper 1 Offline
+3
-1
Let S be the circle in the XY-plane defined the equation x2 + y2 = 4.

Let E1E2 and F1F2 be the chords of S passing through the point P0 (1, 1) and parallel to the X-axis and the Y-axis, respectively. Let G1G2 be the chord of S passing through P0 and having slope$$-$$1. Let the tangents to S at E1 and E2 meet at E3, then tangents to S at F1 and F2 meet at F3, and the tangents to S at G1 and G2 meet at G3. Then, the points E3, F3 and G3 lie on the curve
A
x + y = 4
B
(x $$-$$ 4)2 + (y $$-$$ 4)2 = 16
C
(x $$-$$ 4)(y $$-$$ 4) = 4
D
xy = 4
2
JEE Advanced 2018 Paper 1 Offline
+3
-1
Let S be the circle in the XY-plane defined the equation x2 + y2 = 4.

Let P be a point on the circle S with both coordinates being positive. Let the tangent to S at P intersect the coordinate axes at the points M and N. Then, the mid-point of the line segment MN must lie on the curve
A
(x + y)2 = 3xy
B
x2/3 + y2/3 = 24/3
C
x2 + y2 = 2xy
D
x2 + y2 = x2y2
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) $${x^2} - {a^2}{y^2} = {a^2}$$ $$y = mx + \sqrt {{a^2}{m^2} + 1}$$ $$\left( {{{ - {a^2}m} \over {\sqrt {{a^2}{m^2} - 1} }},\,{{ - 1} \over {\sqrt {{a^2}{m^2} - 1} }}} \right)$$
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)
4
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)
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