• # question_answer The length of the chord of the parabola ${{x}^{2}}=4y$ having equation$x-\sqrt{2}y+4\sqrt{2}=0$ is: A) $3\sqrt{2}$ B) $2\sqrt{11}$ C) $8\sqrt{2}$ D) $6\sqrt{3}$

 $x=\sqrt{2}y-4\sqrt{2}$ ${{x}^{2}}=4y$ Solving we get point of intersection $A(-2\sqrt{2},2),B(4\sqrt{2},8)$ $\therefore$      $AB=\sqrt{{{(6\sqrt{2})}^{2}}+{{6}^{2}}}=6\sqrt{3}.$