Triangle $ABC$ is situated within an ellipse whose major and minor axes have lengths 10 and 8, respectively. Point $A$ is located at one focus, point $B$ is located at an endpoint of the minor axis, and point $C$ is located on the ellipse so that the other focus lies on $\overline{BC}.$ Determine the inradius of triangle $ABC.$

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

ok then what happened.