JEE MAIN - Mathematics (2019 - 10th January Evening Slot - No. 23)
The length of the chord of the parabola x2 $$=$$ 4y having equation x – $$\sqrt 2 y + 4\sqrt 2 = 0$$ is -
$$8\sqrt 2 $$
$$6\sqrt 3 $$
$$3\sqrt 2 $$
$$2\sqrt {11} $$
Explanation
x2 = 4y
x $$-$$ $$\sqrt 2 $$y + 4$$\sqrt 2 $$ = 0
Solving together we get
x2 = 4$$\left( {{{x + 4\sqrt 2 } \over {\sqrt 2 }}} \right)$$
$$\sqrt 2 $$x2 + 4x + 16$$\sqrt 2 $$
$$\sqrt 2 $$x2 $$-$$ 4x $$-$$ 16$$\sqrt 2 $$ = 0
x1 + x2 = 2$$\sqrt 2 $$; x1x2 = $${{ - 16\sqrt 2 } \over {\sqrt 2 }}$$ = $$-$$ 16
Similarly,
($$\sqrt 2 $$y $$-$$ 4$$\sqrt 2 $$)2 = 4y
2y2 + 32 $$-$$ 16y = 4y
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$$\ell $$AB = $$\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $$
$$ = \sqrt {{{\left( {2\sqrt 2 } \right)}^2} + 64 + {{\left( {10} \right)}^2} - 4\left( {16} \right)} $$
$$ = \sqrt {8 + 64 + 100 - 64} $$
$$ = \sqrt {108} = 6\sqrt 3 $$
x $$-$$ $$\sqrt 2 $$y + 4$$\sqrt 2 $$ = 0
Solving together we get
x2 = 4$$\left( {{{x + 4\sqrt 2 } \over {\sqrt 2 }}} \right)$$
$$\sqrt 2 $$x2 + 4x + 16$$\sqrt 2 $$
$$\sqrt 2 $$x2 $$-$$ 4x $$-$$ 16$$\sqrt 2 $$ = 0
x1 + x2 = 2$$\sqrt 2 $$; x1x2 = $${{ - 16\sqrt 2 } \over {\sqrt 2 }}$$ = $$-$$ 16
Similarly,
($$\sqrt 2 $$y $$-$$ 4$$\sqrt 2 $$)2 = 4y
2y2 + 32 $$-$$ 16y = 4y
$$\ell $$AB = $$\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $$
$$ = \sqrt {{{\left( {2\sqrt 2 } \right)}^2} + 64 + {{\left( {10} \right)}^2} - 4\left( {16} \right)} $$
$$ = \sqrt {8 + 64 + 100 - 64} $$
$$ = \sqrt {108} = 6\sqrt 3 $$
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