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JEE Advance - Mathematics (2017 - Paper 1 Offline - No. 13)

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) $${x^2} - {a^2}{y^2} = {a^2}$$ $$y = mx + \sqrt {{a^2}{m^2} + 1} $$ $$\left( {{{ - {a^2}m} \over {\sqrt {{a^2}{m^2} - 1} }},\,{{ - 1} \over {\sqrt {{a^2}{m^2} - 1} }}} \right)$$
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) $${x^2} - {a^2}{y^2} = {a^2}$$ $$y = mx + \sqrt {{a^2}{m^2} + 1} $$ $$\left( {{{ - {a^2}m} \over {\sqrt {{a^2}{m^2} - 1} }},\,{{ - 1} \over {\sqrt {{a^2}{m^2} - 1} }}} \right)$$
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?
(I) (ii) Q)
(I) (ii) (P)
(III) (i) (P)
(II) (ii) (Q)

Explanation

$$\bullet$$ For the given equation $$a = \sqrt 2 $$ and point ($$-$$1, 1), x2 + y2 = a2 satisfies these conditions (i.e. I of Column 1)

$$\bullet$$ Thus, equation of tangent is $$ - x + y = 2 \Rightarrow y = x + 2$$; thus (ii) of Column 2 satisfies this condition with m = 1. That is, $$y = mx + a\sqrt {{m^2} + 1} $$

$$\bullet$$ It is given that point of constant ($$-$$1, 1) given m = 1 and $$a = \sqrt 2 $$ Q; of Columns 3 : $$\left( {{{ - ma} \over {\sqrt {{m^2} + 1} }},{a \over {\sqrt {{m^2} + 1} }}} \right)$$ gives point of contact ($$-$$1, 1).

Thus, correct combination is (I) (II) (Q); thus, option (A) is correct.

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