JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
The length of the chord of the parabola \(x^2 = 4y\) having equation \(x - \sqrt 2y + 4\sqrt 2 = 0\) is
- A \(3\sqrt 2\)
- B \(2\sqrt {11}\)
- C \(8\sqrt 2\)
- D \(6\sqrt 3\)
Answer & Solution
Correct Answer
(D) \(6\sqrt 3\)
Step-by-step Solution
Detailed explanation
\(x = \sqrt 2 y - 4\sqrt 2 \) \({x^2} = 4y\) Solving we get point of intersection \(A\left( { - 2\sqrt 2 ,2} \right),B\left( {4\sqrt 2 ,8} \right)\) \(\therefore AB = \sqrt {{{\left( {6\sqrt 2 } \right)}^2} + {6^2}} = 6\sqrt 3 \)
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