A point moves such that the sum of its distan | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the point be $P(x, y, z)$. The given condition is $PF_1 + PF_2 = 10$,where $F_1 = (4, 0, 0)$ and $F_2 = (-4, 0, 0)$.$\sqrt{(x-4)^2 + y^2 + z^2

Vedclass · Accepted Answer

$9x^2 + 25y^2 + 25z^2 = 225$

Vedclass · Answer

(C) Let the point be $P(x, y, z)$. The given condition is $PF_1 + PF_2 = 10$,where $F_1 = (4, 0, 0)$ and $F_2 = (-4, 0, 0)$.$\sqrt{(x-4)^2 + y^2 + z^2} + \sqrt{(x+4)^2 + y^2 + z^2} = 10$Let $S = x^2 + y^2 + z^2 + 16$. Then the equation is $\sqrt{S - 8x} + \sqrt{S + 8x} = 10$.Squaring both sides: $(S - 8x) + (S + 8x) + 2\sqrt{S^2 - 64x^2} = 100$$2S + 2\sqrt{S^2 - 64x^2} = 100 \Rightarrow S + \sqrt{S^2 - 64x^2} = 50$$\sqrt{S^2 - 64x^2} = 50 - S$Squaring again: $S^2 - 64x^2 = 2500 - 100S + S^2$$100S - 64x^2 = 2500$Substitute $S = x^2 + y^2 + z^2 + 16$:$100(x^2 + y^2 + z^2 + 16) - 64x^2 = 2500$$100x^2 + 100y^2 + 100z^2 + 1600 - 64x^2 = 2500$$36x^2 + 100y^2 + 100z^2 = 900$Dividing by $4$: $9x^2 + 25y^2 + 25z^2 = 225$.

$A$ point moves such that the sum of its distances from the points $(4, 0, 0)$ and $(-4, 0, 0)$ remains $10$. The locus of the point is

Similar Questions

If $A$ and $B$ are the points $(3, 4, 5)$ and $(-1, 3, -7)$ respectively,find the equation of the set of points $P$ such that $PA^{2} + PB^{2} = k^{2}$,where $k$ is a constant.

Find the radius of the circle formed by the intersection of the sphere $x^{2} + y^{2} + z^{2} - 2y - 4z = 11$ and the plane $x + 2y + 2z = 15$.

The equation $|r|^2 - r \cdot (2i + 4j - 2k) - 10 = 0$ represents a

The coordinate of a point equidistant from the points $(0, 0, 0), (a, 0, 0), (0, b, 0),$ and $(0, 0, c)$ is

If $(2, 3, 5)$ is one end of a diameter of the sphere ${x^2} + {y^2} + {z^2} - 6x - 12y - 2z + 20 = 0$,then the coordinates of the other end of the diameter are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module