Question

# The length of the chord of the parabola $$x^2 = 4y$$ having equation $$x - \sqrt{2}y + 4\sqrt{2} = 0$$ is :

A
211
B
32
C
63
D
82

Solution

## The correct option is D $$6\sqrt{3}$$$$x^2 = 4y$$$$x - \sqrt{2} y + 4\sqrt{2} = 0$$Solving together we get$$x^2 = 4 \left(\dfrac{x + 4\sqrt{2}}{\sqrt{2}}\right)$$$$\sqrt{2}x^2 - 4x - 16\sqrt{2}=0$$$$x_1 + x_2 = 2\sqrt{2}$$;  $$x_1x_2 = \dfrac{-16\sqrt{2}}{\sqrt{2}}=-16$$Similarly,$$(\sqrt{2} y - 4\sqrt{2})^2 = 4y$$$$2y^2 + 32 - 16y = 4y$$$$2y^2 -20y + 32 = 0 < ^{y_1 + y_2 = 10}_{y_1y_2=16}$$$$\ell_{AB} = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$$ $$= \sqrt{(2\sqrt{2})^2 + 64 + (10)^2 - 4(16)}$$$$=\sqrt{8 + 64 + 100 - 64}$$$$= \sqrt{108} = 6\sqrt{3}$$Mathematics

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