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Find the equation of the set of points $P$ the sum of whose distance from $(4,0,0)$ and $(-4,0,0)$ is equal to $10$

$\begin{array}{1 1}16x^2+25y^2+25z^2+225=0 \\9x^2+25y^2+25z^2-225=0 \\ 9x^2+25y^2+25z^2+225=0 \\16x^2+25y^2+25z^2-225=0 \end{array} $

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  • The distance between two points $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$ is given by $\sqrt {(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$
Let the given points be $A(4,0,0)$ and $B(-4,0,0)$
Let $ P(x,y,z)$ be the set of points the sum of whose distance from $A$ and $B$ is 10.
$\Rightarrow\:\overline {PA}+\overline {PB}=10$
$\Rightarrow\:\overline {PA}=10-\overline {PB}$
Squaring on both the sides
$(\overline {PA})^2=100+(\overline {PB})^2-20\overline {PB}$........(i)
We know that the distance between two points $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$ is given by $\sqrt {(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$
$(\overline {PA})^2=(x-4)^2+y^2+z^2=x^2+y^2+z^2-8x+16$
$(\overline {PB})^2=(x+4)^2+y^2+z^2=x^2+y^2+z^2+8x+16$
Substituting the values in (i) we get
$\Rightarrow\:x^2+y^2+z^2-8x+16=100+(x^2+y^2+z^2+8x+16)-20.(\sqrt {x^2+y^2+z^2+8x+16})$
$\Rightarrow\:-16x=100-20(\sqrt {x^2+y^2+z^2+8x+16})$
$\Rightarrow\:-4x=25-5(\sqrt {x^2+y^2+z^2+8x+16})$
$\Rightarrow\:-4x-25=-5(\sqrt {x^2+y^2+z^2+8x+16})$
$\Rightarrow\:4x+25=5(\sqrt {x^2+y^2+z^2+8x+16})$
Again squaring on both the sides we get
$16x^2+625+200x=25(x^2+y^2+z^2+8x+16)$
$\Rightarrow\:9x^2+25y^2+25z^2-225=0$
$i.e.,$ The required equation of the set of points $P$ is
$9x^2+25y^2+25z^2-225=0$
answered May 3, 2014 by rvidyagovindarajan_1
 

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