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Parabola question

2017 · Shift 1 · Q33
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  5. /2017 · Shift 1 · Q33

Parabola question

2017 · Shift 1 · Q33

JEE AdvancedMathematicsParabolaMCQ+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) x2+y2=a{x^2} + {y^2} = ax2+y2=a my=m2x+amy = {m^2}x + amy=m2x+a (am2, 2am)\left( {{a \over {{m^2}}},\,{{2a} \over m}} \right)(m2a​,m2a​)
(ii) x2a2y2=a2]{x^2}{a^2}{y^2} = {a^2}]x2a2y2=a2] y=mx+am2+1y = mx + a\sqrt {{m^2} + 1}y=mx+am2+1​ (−mam2+1, am2+1)\left( {{{ - ma} \over {\sqrt {{m^2} + 1} }},\,{a \over {\sqrt {{m^2} + 1} }}} \right)(m2+1​−ma​,m2+1​a​)
(iii) y2=4ax{y^2} = 4axy2=4ax y=mx+a2m2−1y = mx + \sqrt {{a^2}{m^2} - 1}y=mx+a2m2−1​ (−a2ma2m2+1, 1a2m2+1)\left( {{{ - {a^2}m} \over {\sqrt {{a^2}{m^2} + 1} }},\,{1 \over {\sqrt {{a^2}{m^2} + 1} }}} \right)(a2m2+1​−a2m​,a2m2+1​1​)
(iv) x2−a2y2=a2{x^2} - {a^2}{y^2} = {a^2}x2−a2y2=a2 y=mx+a2m2+1y = mx + \sqrt {{a^2}{m^2} + 1}y=mx+a2m2+1​ (−a2ma2m2−1, −1a2m2−1)\left( {{{ - {a^2}m} \over {\sqrt {{a^2}{m^2} - 1} }},\,{{ - 1} \over {\sqrt {{a^2}{m^2} - 1} }}} \right)(a2m2−1​−a2m​,a2m2−1​−1​)
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?
  1. A
    (III) (i) (P)
  2. B
    (I) (ii) (Q)
  3. C
    (II) (iv) (R)
  4. D
    (III) (ii) (Q)
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