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# Find the equation of locus of a point, where the sum of whose distances is 6 from (0, 2) and (0, -2).

Last updated date: 24th Feb 2024
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Hint: A locus is a curve or other shape made by all the points satisfying a particular equation of the relation between the coordinates, or by a point, line, or moving surface. All the shapes such as circle, ellipse, parabola, hyperbola, are defined by the locus as a set of points.
Formula used:
$d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_2}} \right)}^2}}$

Complete step-by-step solution:
According to the question, there is a point (Let us suppose that is O), and there are two other points that are A and B with coordinates (0,2) and (0,-2) respectively.
Somewhat like this:

As per the question, $\left( {OA} \right) + \left( {OB} \right) = 6$, so we have to find out OA and OB. We know that when 2 points are given, we will find the distance in between two points by the formula:
$d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_2}} \right)}^2}}$
Now, by Calculating OA and OB, we get:
$OA = \sqrt {{{\left( {{x_1} - 0} \right)}^2} + {{\left( {{y_1} - 2} \right)}^2}}$
This can be further simplified as:
$OA = \sqrt {x_1^2 + {{\left( {{y_1} - 2} \right)}^2}}$
$OB = \sqrt {x_1^2 + {{\left( {{y_1} + 2} \right)}^2}}$
Now, by putting these values of OA and OB in the formula of d above as:
$\sqrt {x_1^2 + {{\left( {{y_1} - 2} \right)}^2}} + \sqrt {x_1^2 + {{\left( {{y_1} + 2} \right)}^2}} = 6$
Now, by using the identity of ${\left( {a - b} \right)^2}$, we get:
$\sqrt {x_1^2 + y_1^2 + 4 - 4{y_1}} + \sqrt {x_1^2 + {y_1} + 4 + 4{y_1}} = 6$
By reshuffling or rearranging this equation, we get:
$\left( {\sqrt {x_1^2 + y_1^2 + 4 - 4{y_1}} - 6} \right) = \left( {\sqrt {x_1^2 + {y_1} + 4 + 4{y_1}} } \right)$
Now, by squaring both the sides, we get:
$x_1^2 + y_1^2 + 4 - 4{y_1} + 36 - 12\sqrt {x_1^2 + y_1^2 + 4 - 4{y_1}} = x_1^2 + y_1^2 + 4 + 4{y_1}$
After cancelling terms, we get:
$\Rightarrow - 8{y_1} + 36 = 12\sqrt {x_1^2 + y_1^2 + 4 - 4{y_1}} \\ \Rightarrow - \dfrac{8}{{12}}{y_1} + \dfrac{{36}}{{12}} = \sqrt {x_1^2 + y_1^2 + 4 - 4{y_2}} \\ \Rightarrow - \dfrac{2}{3}{y_1} + 3 = \sqrt {{x_1}^2 + y_1^2 + 4 - 4{y_1}} \\$
Squaring both the sides, and putting the identity of ${\left( {a - b} \right)^2}$, we get:
$\Rightarrow \dfrac{4}{9}y_1^2 + 9 - 4{y_1} = x_1^2 + y_1^2 + 4 - 4{y_1}$
By cancelling$4{y_1}$and reshuffling, we get:
$\Rightarrow 0 = x_1^2 + y_1^2 - \dfrac{4}{9}y_1^2 + 4 - 9$
$\Rightarrow x_1^2 + y_1^2\dfrac{5}{9} = 5 \\ \Rightarrow \dfrac{{x_1^2}}{5} + \dfrac{{y_1^2}}{5} \times \dfrac{5}{9} = 1 \\ \Rightarrow \dfrac{{x_1^2}}{5} + \dfrac{{y_1^2}}{9} = 1 \\ \Rightarrow \left( {\dfrac{{{x^2}}}{5} + \dfrac{{{y^2}}}{9} = 1} \right)$
This is an equation of ellipse, comparing with general equation of ellipse:
$\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ we get,
$a = \sqrt 5 \\ b = \sqrt 9 = 3$

Note: An ellipse if we speak in terms of locus, it is the set of all points on an XY-plane, whose distance from two fixed points (known as foci) adds up to a constant value. When the centre of the ellipse is at the origin (0,0) and the foci are on the x-axis and y-axis, then equation is $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$.