Triangle $ABC$ is situated within an ellipse whose major and minor axes have lengths 10 and 8, respectively. Point $A$ is locate
1 answer:
Answer:
The coordinates of the incenter ≈ (0.0563, 0.136)
The length of the inradius ≈ 2.3634
Step-by-step explanation:
The lengths of the major and minor axis of the ellipse in which ΔABC is situated are;
Major axis = 10, minor axis = 8
The location of point A = The focus of the ellipse
The location of point B = An endpoint of the minor axis
The location of point C = On the ellipse such that the other focus lies on
B(0, 4), P(3, 0)
The equation of the line BPC y - 0 = ((-4)/3)·(x - 3)
y = -4·x/3 + 4
Therefore, at 'C', we have;
1 = x²/5² + y²/4² = x²/5² + (-4·x/3 + 4)²/4²
Using an online tool, we get;
1184·x²/49 = 400
x = ± √(400×49/1184)
At C, x = √(400×49/1184) ≈ 4.07
y = -4 × (√(400×49/1184))/3 + 4 ≈ -1.425
The coordinates of point C = (4.07, -1.425)
The coordinates of point A = (-3, 0)
The coordinates of point B = (0, 4)
The length of AB = √(4² + (0 - 3)²) = 5
The length of AC = √((-3 - 4.07)² + (0 - (-1.425))²) ≈ 7.2122
The length of BC = √((0 - 4.07)² + (4 - (-1.425))²) ≈ 6.782
The coordinates of the incenter = (4.07 + (-3) + 0)/(5 + 7.2122 + 6.782), (0 + 4 + (-1.425))/(5 + 7.2122 + 6.782)) ≈ (0.0563, 0.136)
The perpendicular to the line AB (y = (4/3)·x + 4) from the incenter
The slope equation of the line perpendicular to AB = -3/4
The equation of the line perpendicular to AB = y - 0.136 = (-3/4)·(x - 0.0563)
y = (-3/4)·(x - 0.0563) + 0.136
∴ At the point of intersection, (-3/4)·(x - 0.0563) + 0.136 = (4/3)·x + 4)
Solving gives, x = -(46.368 - 2.0268/4)/25 ≈ -1.83442
y = (-3/4)×((-1.83442) - 0.0563) + 0.136 ≈ 1.55404
The radius ≈ √((1.55404 - 0.136)² + ((-1.83442) - 0.0563)²) ≈ 2.3634
The length of the inradius ≈ 2.3634
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