Find the equation of the locus of a point p which is equidistant from the y axis and the point (3,-1) ?
i got x^2-6x+y^2+2y+10 as the answer not sure if it is right
i got x^2-6x+y^2+2y+10 as the answer not sure if it is right
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First of all, your answer is not even an equation since the equals sign is missing. Second of all, the locus of points equidistant from a line (directrix) and a point (focus) is a parabola, not a circle; thus only one of the variables should be squared.
There are at least two methods. The first method involves using a directrix and a focus; the second method just involves directly setting the two requested distances equal.
Method 1: We treat the y-axis as the directrix and the point (3,-1) as the focus. Along the line through the focus that is perpendicular to the directrix, the vertex is halfway from the focus to the directrix. So the vertex is halfway from (3,-1) to (0,-1), which is (3/2,-1).
The parabola opens away from the directrix, and the axis of symmetry of the parabola is perpendicular to the directrix. So in this case, the parabola opens sideways to the right. The formula becomes
(y - y_0)^2 = 4a(x - x_0)
where (x_0, y_0) is the vertex, and a is the distance from the vertex to the focus.
Here, x_0 = 3/2, y_0 = -1, and a = 3 - 3/2 = 3/2.
The equation of the locus becomes
(y + 1)^2 = 4(3/2)(x - 3/2)
(y + 1)^2 = 6x - 9
-6x + y^2 + 2y + 10 = 0.
As you can see, your answer was close but not quite correct. There should be no x^2 term, and the expression needs to be set equal to 0.
Method 2: Set the distance from (x,y) to the y-axis equal to the distance from (3,-1) to (x,y).
Then |x| = sqrt((x-3)^2 + (y+1)^2)
x = +- sqrt((x-3)^2 + (y+1)^2)
x^2 = (x-3)^2 + (y+1)^2 = x^2 - 6x + 9 + y^2 + 2y + 1
Finally, once again we have -6x + y^2 + 2y + 10 = 0.
Lord bless you today!
There are at least two methods. The first method involves using a directrix and a focus; the second method just involves directly setting the two requested distances equal.
Method 1: We treat the y-axis as the directrix and the point (3,-1) as the focus. Along the line through the focus that is perpendicular to the directrix, the vertex is halfway from the focus to the directrix. So the vertex is halfway from (3,-1) to (0,-1), which is (3/2,-1).
The parabola opens away from the directrix, and the axis of symmetry of the parabola is perpendicular to the directrix. So in this case, the parabola opens sideways to the right. The formula becomes
(y - y_0)^2 = 4a(x - x_0)
where (x_0, y_0) is the vertex, and a is the distance from the vertex to the focus.
Here, x_0 = 3/2, y_0 = -1, and a = 3 - 3/2 = 3/2.
The equation of the locus becomes
(y + 1)^2 = 4(3/2)(x - 3/2)
(y + 1)^2 = 6x - 9
-6x + y^2 + 2y + 10 = 0.
As you can see, your answer was close but not quite correct. There should be no x^2 term, and the expression needs to be set equal to 0.
Method 2: Set the distance from (x,y) to the y-axis equal to the distance from (3,-1) to (x,y).
Then |x| = sqrt((x-3)^2 + (y+1)^2)
x = +- sqrt((x-3)^2 + (y+1)^2)
x^2 = (x-3)^2 + (y+1)^2 = x^2 - 6x + 9 + y^2 + 2y + 1
Finally, once again we have -6x + y^2 + 2y + 10 = 0.
Lord bless you today!
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Hi,
All the points equidistant from a line and a point not on the line form a parabola. In this case, it opens to the right, so its formula would be x = a(y - k)² + h with a vertex (h,k).
The vertex would be (h,k) = (1.5, -1).
x = a(y + 1)² + 1.5
The distance from the vertex to the focus is 1.5, so p = 1.5.
Since a = 1/(4p), then a = 1/(4*1.5), so a = 1/6
x = 1/6(y + 1)² + 1.5
6x = (y + 1)² + 9
6x = y² + 2y + 1 + 9
y² - 6x + 2y + 10 = 0 <==ANSWER
I hope that helps!! :-)
All the points equidistant from a line and a point not on the line form a parabola. In this case, it opens to the right, so its formula would be x = a(y - k)² + h with a vertex (h,k).
The vertex would be (h,k) = (1.5, -1).
x = a(y + 1)² + 1.5
The distance from the vertex to the focus is 1.5, so p = 1.5.
Since a = 1/(4p), then a = 1/(4*1.5), so a = 1/6
x = 1/6(y + 1)² + 1.5
6x = (y + 1)² + 9
6x = y² + 2y + 1 + 9
y² - 6x + 2y + 10 = 0 <==ANSWER
I hope that helps!! :-)