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JEE Advance - Mathematics (2017 - Paper 1 Offline - No. 15)

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) $${x^2} - {a^2}{y^2} = {a^2}$$ $$y = mx + \sqrt {{a^2}{m^2} + 1} $$ $$\left( {{{ - {a^2}m} \over {\sqrt {{a^2}{m^2} - 1} }},\,{{ - 1} \over {\sqrt {{a^2}{m^2} - 1} }}} \right)$$
If a tangent to a suitable conic (Column 1) is found to be y = x + 8 and its point of contact is (8, 16), then which of the following options is the only CORRECT combination?
(III) (i) (P)
(I) (ii) (Q)
(II) (iv) (R)
(III) (ii) (Q)

Explicació

The tangent is y = x + 8.

Satisfying (8, 16) in y2 = 4ax, we have

162 = 4a . 8. $$\therefore$$ a = 8

On comparing y = mx + $${a \over m}$$, we have m = 1

The point of contact is $$\left( {{a \over {{m^2}}},{{2a} \over m}} \right) = (8,16)$$

So its verified.

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