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The locus of a variable point whose distance from (−2,0) is 2/3 times its distance from the line x=−9/2 is?
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The locus of a variable point whose distance from (−2,0) is 2/3 times ...
By definition , it is an ellipse.
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The locus of a variable point whose distance from (−2,0) is 2/3 times ...
Solution:
To determine the locus of a variable point, we need to find the set of all points that satisfy a given condition. In this case, we want to find the locus of a point that satisfies two conditions: its distance from (-2,0) is 2/3 times its distance from the line x=-9/2.
Condition 1: Distance from (-2,0)
Let's start by considering the first condition, which states that the distance from the point to (-2,0) is a certain value. We can represent this distance as d1.
Using the distance formula, the distance d1 between any point (x,y) and (-2,0) can be found as:
d1 = sqrt((x - (-2))^2 + (y - 0)^2)
= sqrt((x + 2)^2 + y^2)
Condition 2: Distance from the line x = -9/2
Now, let's consider the second condition, which states that the distance from the point to the line x = -9/2 is a certain value. We can represent this distance as d2.
The distance d2 between a point (x,y) and the line x = -9/2 is the perpendicular distance from the point to the line. Since the line is vertical, the distance is simply the absolute value of the difference between the x-coordinates of the point and the line, which is:
d2 = |x - (-9/2)|
= |x + 9/2|
Locus of the Variable Point:
Now, we need to find the set of all points that satisfy both conditions simultaneously. In other words, we need to find the locus of the point.
According to the given condition, the distance from (-2,0) is 2/3 times the distance from the line x = -9/2. Mathematically, this can be expressed as:
d1 = (2/3) * d2
Substituting the expressions for d1 and d2, we get:
sqrt((x + 2)^2 + y^2) = (2/3) * |x + 9/2|
To simplify this equation, we can square both sides:
(x + 2)^2 + y^2 = (2/3)^2 * (x + 9/2)^2
(x + 2)^2 + y^2 = (4/9) * (x + 9/2)^2
Expanding and simplifying the equation, we get:
x^2 + 4x + 4 + y^2 = (4/9) * (x^2 + 9x + (9/2)^2)
x^2 + 4x + 4 + y^2 = (4/9) * (x^2 + 9x + 81/4)
Multiplying both sides by 9 to eliminate the fraction, we get:
9x^2 + 36x + 36 + 9y^2 = 4x^2 + 36x + 81
Simplifying further, we have:
9x^2 + 9y^2 = 4x^2 + 45
Rearranging the terms, we get
To determine the locus of a variable point, we need to find the set of all points that satisfy a given condition. In this case, we want to find the locus of a point that satisfies two conditions: its distance from (-2,0) is 2/3 times its distance from the line x=-9/2.
Condition 1: Distance from (-2,0)
Let's start by considering the first condition, which states that the distance from the point to (-2,0) is a certain value. We can represent this distance as d1.
Using the distance formula, the distance d1 between any point (x,y) and (-2,0) can be found as:
d1 = sqrt((x - (-2))^2 + (y - 0)^2)
= sqrt((x + 2)^2 + y^2)
Condition 2: Distance from the line x = -9/2
Now, let's consider the second condition, which states that the distance from the point to the line x = -9/2 is a certain value. We can represent this distance as d2.
The distance d2 between a point (x,y) and the line x = -9/2 is the perpendicular distance from the point to the line. Since the line is vertical, the distance is simply the absolute value of the difference between the x-coordinates of the point and the line, which is:
d2 = |x - (-9/2)|
= |x + 9/2|
Locus of the Variable Point:
Now, we need to find the set of all points that satisfy both conditions simultaneously. In other words, we need to find the locus of the point.
According to the given condition, the distance from (-2,0) is 2/3 times the distance from the line x = -9/2. Mathematically, this can be expressed as:
d1 = (2/3) * d2
Substituting the expressions for d1 and d2, we get:
sqrt((x + 2)^2 + y^2) = (2/3) * |x + 9/2|
To simplify this equation, we can square both sides:
(x + 2)^2 + y^2 = (2/3)^2 * (x + 9/2)^2
(x + 2)^2 + y^2 = (4/9) * (x + 9/2)^2
Expanding and simplifying the equation, we get:
x^2 + 4x + 4 + y^2 = (4/9) * (x^2 + 9x + (9/2)^2)
x^2 + 4x + 4 + y^2 = (4/9) * (x^2 + 9x + 81/4)
Multiplying both sides by 9 to eliminate the fraction, we get:
9x^2 + 36x + 36 + 9y^2 = 4x^2 + 36x + 81
Simplifying further, we have:
9x^2 + 9y^2 = 4x^2 + 45
Rearranging the terms, we get
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The locus of a variable point whose distance from (−2,0) is 2/3 times its distance from the line x=−9/2 is?
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The locus of a variable point whose distance from (−2,0) is 2/3 times its distance from the line x=−9/2 is? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about The locus of a variable point whose distance from (−2,0) is 2/3 times its distance from the line x=−9/2 is? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The locus of a variable point whose distance from (−2,0) is 2/3 times its distance from the line x=−9/2 is?.
The locus of a variable point whose distance from (−2,0) is 2/3 times its distance from the line x=−9/2 is? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about The locus of a variable point whose distance from (−2,0) is 2/3 times its distance from the line x=−9/2 is? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The locus of a variable point whose distance from (−2,0) is 2/3 times its distance from the line x=−9/2 is?.
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