Derive the equation of the locus of all points that are equidistant from the point F(-3,4) and the line y=-2. Leave your answer in general form. [General Form: ax^2+by^2+cx+dy+e=0]
It will be a parabola. The vertical axis will be at x = -3. F is the focal point. The vertex of the parabola is equidistant from F and the y = -2 line, at x = -3, y = 1.
y = a (x+3)^2 + 1 is the equation.
Do some reading up on parabolas to figure out what a is. I think you will find it is 4 times the distance from the focus to the vertex, or 4*3 = 12
That would make the equation
y = 12(x+3)^2 + 1
Expand and rearrange that to the "general form".
To derive the equation of the locus of all points equidistant from a fixed point and a given line, we can use the method of distance formula.
Let's start by finding the equation of the locus. We know that the locus will be equidistant from the point F(-3, 4) and the line y = -2.
Step 1: Distance from a point to a line
The distance between a point (x1, y1) and a line Ax + By + C = 0 is given by the formula:
d = |Ax1 + By1 + C| / √(A^2 +B^2)
Step 2: Distance from a point to a fixed point
The distance between two points (x1, y1) and (x2, y2) is given by the formula:
d = √((x2 - x1)^2 + (y2 - y1)^2)
Using these two formulas, let's proceed to find the locus.
Step 3: Distance from a point to the line y = -2
We substitute the values from the given line equation into the distance formula mentioned in Step 1:
d = |Ax + B(-2) + C| / √(A^2 + B^2)
= |Ax - 2B + C| / √(A^2 + B^2)
Step 4: Distance from a point to the point F(-3, 4)
We substitute the values from the given point into the distance formula mentioned in Step 2:
d = √((x - (-3))^2 + (y - 4)^2)
= √((x + 3)^2 + (y - 4)^2)
Step 5: Equating the distances
Since the locus is equidistant from the point F(-3, 4) and the line y = -2, we equate the distances found in Step 3 and Step 4:
|Ax - 2B + C| / √(A^2 + B^2) = √((x + 3)^2 + (y - 4)^2)
Step 6: Squaring both sides and eliminating the square root
By squaring both sides of the equation, we can eliminate the square root:
(Ax - 2B + C)^2 / (A^2 + B^2) = (x + 3)^2 + (y - 4)^2
Step 7: Expanding the equation and bringing all terms to one side
Expanding the equation, we get:
(A^2)(x^2) - 2ABx + 2ACx - 4B^2 + 4BC - C^2 = x^2 + 6x + 9 + y^2 - 8y + 16
Rearranging the terms, we have:
(A^2 - 1)x^2 + (2AC - 2AB - 6)x + (4BC - 4B^2 + 8y - 8 - C^2 - 9 - 16) = 0
Finally, we can express the equation in general form, which is ax^2 + by^2 + cx + dy + e = 0:
(A^2 - 1)x^2 + 2(AC - AB - 3)x + 4(y - 2B) + 15 - C^2 = 0
Therefore, the equation of the locus of all points equidistant from F(-3, 4) and the line y = -2 is:
(A^2 - 1)x^2 + 2(AC - AB - 3)x + 4(y - 2B) + 15 - C^2 = 0