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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).

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Last updated date: 24th Feb 2024
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IVSAT 2024
Answer
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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:
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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 \[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\].

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