In the parabola $y^{2} - 2 y + 8 x - 23 = 0$ the length of double ordinate at a distance $4$ units from its vertex is

1

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8

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12

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8 \sqrt{2}

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Solution

The correct option is D $8 \sqrt{2}$
Solution :
$y^{2} - 2 y + 8 x - 23 = 0$ .
$y^{2} - 2 y + 1 = - 8 x + 24$
$(y - 1)^{2} = - 8 (x - 3)$ vertex is $(3, 1)$
Double ordinate with cutting point on
axis $= (3 - 4, 1) = (- 1, 1)$
Put $x = - 1$
$(y - 1)^{2} = 32$
$y = 4 \sqrt{2} + 1 - 4 \sqrt{2} + 1$
Length $= 4 \sqrt{2} + 1 - (- 4 \sqrt{2} + 1)$
$= 8 \sqrt{2}$ .
Dis correct

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In the parabola y2−2y+8x−23=0 the length of double ordinate at a distance 4 units from its vertex is

The correct option is D 8√2Solution :y2−2y+8x−23=0.y2−2y+1=−8x+24(y−1)2=−8(x−3) vertex is (3,1)Double ordinate with cutting point on axis =(3−4,1)=(−1,1)Put x=−1(y−1)2=32y=4√2+1−4√2+1Length =4√2+1−(−4√2+1)=8√2.Dis correct

In the parabola $y^{2} - 2 y + 8 x - 23 = 0$ the length of double ordinate at a distance $4$ units from its vertex is