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Question

Question: The equation of the locus of a point whose distance from (a, 0) is equal to its distance from y-axis...

The equation of the locus of a point whose distance from (a, 0) is equal to its distance from y-axis, is.

A

y22ax=a2y^{2} - 2ax = a^{2}

B

y22ax+a2=0y^{2} - 2ax + a^{2} = 0

C

y2+2ax+a2=0y^{2} + 2ax + a^{2} = 0

D

y2+2ax=a2y^{2} + 2ax = a^{2}

Answer

y22ax+a2=0y^{2} - 2ax + a^{2} = 0

Explanation

Solution

Let the point be (h,k)( h , k )

So,(ha)2+(k0)2=h2( h - a ) ^ { 2 } + ( k - 0 ) ^ { 2 } = h ^ { 2 } h2+a22ah+k2=h2\Rightarrow h ^ { 2 } + a ^ { 2 } - 2 a h + k ^ { 2 } = h ^ { 2 }

Hence locus is y22ax+a2=0y ^ { 2 } - 2 a x + a ^ { 2 } = 0.