The locus of point whose distance from the point (0,5) is always 3/4 of its distance from line y=8 is an ellipse. Find major and minor axis
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To find the major and minor axes of the ellipse, we first need to understand its properties and equation.
The given locus of points consists of all the points that are equidistant from the point (0,5) and the line y=8. Let's denote a point on the ellipse as (x, y).
Step 1: Distance from the point (0,5)
The distance between the point (x, y) and (0,5) can be found using the distance formula:
d1 = √((x - 0)^2 + (y - 5)^2)
= √(x^2 + (y-5)^2)
Step 2: Distance from the line y=8
The distance between the point (x, y) and the line y=8 can be found using the formula:
d2 = |y - 8|
Step 3: Equating the distances
We are given that the distance from the point (0,5) is always 3/4 of the distance from the line y=8. Mathematically, we can represent this as:
d1 = (3/4) * d2
√(x^2 + (y-5)^2) = (3/4) * |y - 8|
To eliminate the absolute value, we can square both sides of the equation:
x^2 + (y-5)^2 = (9/16) * (y - 8)^2
Step 4: Rearranging the equation
Expanding and simplifying the equation, we get:
x^2 + y^2 - 10y + 25 = (9/16)(y^2 - 16y + 64)
16x^2 + 16y^2 - 160y + 400 = 9y^2 - 144y + 576
Expand further and reorganize:
16x^2 + 7y^2 - 16y - 176y + 176 = 0
16x^2 + 7y^2 - 192y + 176 = 0
Step 5: Standard form of the ellipse equation
To determine the major and minor axes, we need to express the equation in the standard form of an ellipse equation:
A(x-h)^2 + B(y-k)^2 = 1
In our case, we have:
16x^2 + 7y^2 - 192y + 176 = 0
Divide both sides of the equation by 176:
(x^2)/(11) + (y^2)/((16/11) - (48/11)y = 1
Comparing the equation with the standard form, we find:
A = 11, B = (16/11) - (48/11) = -32/11
Therefore, the major axis is along the x-axis with a length of 2√A, and the minor axis is along the y-axis with a length of 2√|B|.
Thus, the major axis of the ellipse is 2√11, and the minor axis is 2√(32/11).