√(x−3)2+(y−2)2+√(x−1)2+(y+2)2=10
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√(x−3)2+(y−2)2+√(x−1)2+(y+2)2=10
y^2 = -x + 3
(y - 0)^2 = 4 (-1/4) (x - 3)
Given,
√(x−3)2+(y−2)2+√(x−1)2+(y+2)2=10
(x - 3) + (y - 2)^2 + (x - 1) + (y + 2)^2 = 10
(x - 3 + x - 1) + (y^2 + 4 - 4y) + (y^2 + 4 + 4y) = 10
2x - 4 + 2y^2 + 8 = 10
2y^2 + 2x = 10 - 8 + 4
2y^2 + 2x = 6
y^2 + x = 3
y^2 = -x + 3
⇒ (y - 0)^2 = 4 (-1/4) (x - 3)
Parabola with the vertex (h, k) = (3, 0)
The focal length |p| = 1/4
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