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QJEEM19Ma1034: JEE Main 2019 Mathematics - Class 11 JEEM19Ma1034 Solution: The length of the chord of the parabola x 2 4y having equation…

The length of the chord of the parabola x2 == 4y having equation x – 2y+42=0\sqrt 2 y + 4\sqrt 2 = 0   is -

Multiple Choice Options

  • A 828\sqrt 2
  • B 636\sqrt 3
  • C 323\sqrt 2
  • D 2112\sqrt {11}

Step-by-Step Verified Solution

x2 = 4y

x −- 2\sqrt 2 y + 4 2\sqrt 2 = 0

Solving together we get

x2 = 4 (x+422)\left( {{{x + 4\sqrt 2 } \over {\sqrt 2 }}} \right)

2\sqrt 2 x2 + 4x + 16 2\sqrt 2

2\sqrt 2 x2 −- 4x −- 16 2\sqrt 2 = 0

x1 + x2 = 2 2\sqrt 2 ; x1x2 = −1622{{ - 16\sqrt 2 } \over {\sqrt 2 }} = −- 16

Similarly,

( 2\sqrt 2 y −- 4 2\sqrt 2 )2 = 4y

2y2 + 32 −- 16y = 4y

JEE Main 2019 (Online) 10th January Evening Slot Mathematics - Parabola Question 106 English Explanation 1
JEE Main 2019 (Online) 10th January Evening Slot Mathematics - Parabola Question 106 English Explanation 2
ℓ\ell AB = (x2−x1)2+(y2−y1)2\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}

=(22)2+64+(10)2−4(16)= \sqrt {{{\left( {2\sqrt 2 } \right)}^2} + 64 + {{\left( {10} \right)}^2} - 4\left( {16} \right)}

=8+64+100−64= \sqrt {8 + 64 + 100 - 64}

=108=63= \sqrt {108} = 6\sqrt 3

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