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JEE Advance - Mathematics (2017 - Paper 1 Offline - No. 14)

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) $${x^2} - {a^2}{y^2} = {a^2}$$ $$y = mx + \sqrt {{a^2}{m^2} + 1} $$ $$\left( {{{ - {a^2}m} \over {\sqrt {{a^2}{m^2} - 1} }},\,{{ - 1} \over {\sqrt {{a^2}{m^2} - 1} }}} \right)$$
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?
(IV) (iv) (S)
(II) (iv) (R)
(IV) (iii) (S)
(II) (ii) (R)

Explanation

Given, tangent to conic is $$\sqrt 3 x + 2y = 4$$ and point of contact is

$$\left( {\sqrt 3 ,{1 \over 2}} \right)$$ ..... (1)

$$\sqrt 3 x + 2y = 4 \Rightarrow 2y = 4 - \sqrt 3 x$$

$$ \Rightarrow y = {{ - \sqrt 3 } \over 2}x + 2$$ ....... (2)

On comparing this equation with options in Column 2, we get

$$m = {{ - \sqrt 3 } \over 2}$$

Equation (1) can also be written as

$$ \Rightarrow \sqrt 3 x + 4\left( {{1 \over 2}} \right)y = 4$$

On comparing with the option (II), we get

$${a^2} = 4 \Rightarrow a = 2$$

Now, for a = 2, $$m = {{ - \sqrt 3 } \over 2}\left( {{{ - {a^2}m} \over {\sqrt {{a^2}{m^2} + 1} }},{1 \over {\sqrt {{a^2}{m^2} + 1} }}} \right)$$ gives $$\left( {\sqrt 3 ,{1 \over 2}} \right)$$ as the point of contact.

Thus, the correct combination is (II) (iv) (R); hence, option (B) is correct.

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