Let A(2,3) be a fixed point. A point P moves such that PA is equal to the distance of P from the y-axis. Find the equation of the locus of P
PA2 = (x-x1)^2 + (y-y1)^2 +???
= (x-2)^2 + (y-3)^2 +???
I tried solving it using the distance formula but I'm still confused because I am given only one point A (2,3) coordinate. I searched all blogs on google but couldn't find a solved problem similar to this. Please help!
Ths distance of (x,y) from the y-axis is just x. So,
√((x-2)^2 + (y-3)^2) = x
(x-2)^2 + (y-3)^2 = x^2
x^2-4x+4 + (y-3)^2 = x^2
(y-3)^2 = 4x-4
(y-3)^2 = 4(x-1)
as I showed using the properties of parabolas
Well, fret not! Let's try to find the equation of the locus of point P step by step.
Given that point A(2,3) is fixed, and point P moves such that PA is equal to the distance of P from the y-axis.
Now, let point P be represented by coordinates (x, y). We know that PA is equal to the distance of P from the y-axis. In other words, PA = |x|.
Using the distance formula, we have:
PA² = (x - 2)² + (y - 3)²
Since PA = |x|, we can substitute |x| for PA:
|x|² = (x - 2)² + (y - 3)²
Now, let's consider the two cases:
1) When x is positive, |x| = x:
x² = (x - 2)² + (y - 3)²
2) When x is negative, |x| = -x:
(-x)² = (x - 2)² + (y - 3)²
x² = (x - 2)² + (y - 3)²
Combining the two cases, we can simplify the equation:
x² = (x - 2)² + (y - 3)²
Expanding the squared terms, we have:
x² = x² - 4x + 4 + (y - 3)²
0 = -4x + 4 + (y - 3)²
Simplifying further:
4x = (y - 3)² - 4
And finally, rearranging the equation to isolate y, we get:
(y - 3)² = 4x + 4
Thus, the equation of the locus of point P is:
(y - 3)² = 4x + 4
So, it turns out the locus of point P is a parabola!
I hope that helps, or at least brings a smile to your face! Let me know if you have any more questions.
To find the equation of the locus of point P, where PA is equal to the distance of P from the y-axis, we need to use the definition of distance and the given point A(2,3).
Let the coordinates of P be (x, y). The distance of P from the y-axis is given by |x|. The distance formula between two points is:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
In this case, using the distance formula, we have:
PA = sqrt((x - 2)^2 + (y - 3)^2)
We are given that PA is equal to the distance of P from the y-axis, which is |x|:
PA = sqrt((x - 2)^2 + (y - 3)^2) = |x|
To eliminate the absolute value, we can square both sides of the equation:
(sqrt((x - 2)^2 + (y - 3)^2))^2 = (|x|)^2
Simplifying, we get:
(x - 2)^2 + (y - 3)^2 = x^2
Expand and simplify further:
x^2 - 4x + 4 + y^2 - 6y + 9 = x^2
Combine like terms:
-4x + 13 + y^2 - 6y = 0
Rearrange the terms:
y^2 - 6y - 4x + 13 = 0
This is the equation of the locus of P.
To find the equation of the locus of point P, we first need to understand the given conditions. The problem states that point P moves such that the distance between P and point A(2,3) is equal to the distance between P and the y-axis.
Let the coordinates of point P be (x, y). Now, we need to find the distance between point P and point A, which can be calculated using the distance formula:
PA = √((x - x1)^2 + (y - y1)^2)
Substituting the values of point A (2,3):
PA = √((x - 2)^2 + (y - 3)^2)
Now, we need to find the distance between point P and the y-axis. Since the y-axis is a vertical line passing through x = 0, the distance between P and the y-axis is simply the absolute value of the x-coordinate of P:
Distance from P to y-axis = |x|
Now, combining the two distances, we have:
PA = Distance from P to y-axis
√((x - 2)^2 + (y - 3)^2) = |x|
To simplify this equation, we square both sides to remove the square root:
(x - 2)^2 + (y - 3)^2 = x^2
Expanding the equation:
x^2 - 4x + 4 + y^2 - 6y + 9 = x^2
Simplifying further:
- 4x + 4 + y^2 - 6y + 9 = 0
Combine like terms:
y^2 - 6y - 4x + 13 = 0
This is the equation of the locus of point P that satisfies the given conditions.