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Exercise 6.5 · Q1

Q.The equation of the locus of the point whose distance from the yy-axis is half the distance from the origin is

(1) x2+3y2=0x^2+3y^2=0
(2) x2−3y2=0x^2-3y^2=0
(3) 3x2+y2=03x^2+y^2=0
(4) 3x2−y2=03x^2-y^2=0
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✓ Free question

Distance from yy-axis is ∣x∣|x|; distance from origin is x2+y2\sqrt{x^2+y^2}. Set ∣x∣=12x2+y2|x|=\tfrac12\sqrt{x^2+y^2} and square to eliminate the root.

Let P(x,y)P(x,y) be a point on the locus.

Step 1. Write the two distances. The perpendicular distance of PP from the yy-axis is ∣x∣|x|. The distance of PP from the origin O(0,0)O(0,0) is x2+y2\sqrt{x^2+y^2}.

Step 2. Translate the given condition. "Distance from the yy-axis is half the distance from the origin" gives

∣x∣=12x2+y2.|x| = \frac12\sqrt{x^2+y^2}.

Step 3. Square both sides (valid since both sides are non-negative):

x2=14(x2+y2).x^2 = \frac14\left(x^2+y^2\right).

Step 4. Simplify. Multiply by 44: 4x2=x2+y24x^2 = x^2+y^2, so

3x2−y2=0.3x^2 - y^2 = 0.

This matches option (4). Option (3) 3x2+y2=03x^2+y^2=0 is the sign-error version from mishandling the transposition; options (1) and (2) come from squaring the ratio backwards (distance from origin = half distance from yy-axis).

✓Final answer

Option (4): 3x2−y2=03x^2-y^2=0

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