Question

if the sum of the distance of a point from the origin and from the line x=2 is always equal to 4 then locus of this point is

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

GIVEN

The sum of the distance of a point from the origin and from the line x=2 is always equal to 4

TO DETERMINE

The locus of the point

CALCULATION

Let ( h, k ) be the point

Then the distance of the point ( h, k ) from

origin ( 0, 0) is

$= \sf{ \sqrt{ {(h - 0)}^{2} + {(k - 0)}^{2} } \: \: }$

$= \sf{ \sqrt{ {h }^{2} + {k }^{2} } \: \: }$

Again the distance of the point ( h, k ) from the line

x = 2 is h - 2

Hence by the given condition

$\sf{ \sqrt{ {h }^{2} + {k }^{2} } \: + (h - 2) = 4 \: }$

$\implies \: \sf{ \sqrt{ {h }^{2} + {k }^{2} } \: = 6 - h \: }$

Squaring both sides we get

$\implies \: \sf{ {h }^{2} + {k }^{2} \: = {(6 - h)}^{2} \: }$

$\implies \: \sf{ {h }^{2} + {k }^{2} \: = 36 - 12h + {h}^{2} \: }$

$\implies \: \sf{ {k }^{2} \: = 36 - 12h \: }$

Hence the locus of the point is given by

$\: \: \: \sf{ {y }^{2} \: = 36 - 12x \: }$

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