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Home  >>  CBSE XI  >>  Math  >>  Conic Sections

Show that the set of all points such that the difference of their distances from $(4, 0)$ and $(-4, 0)$ is always equal to $2$ represent a hyperbola.

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  • Distance between two points $A(x_1,y_1)$ and $B(x_2,y_2)$ is $AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.
It is given that the difference of the distance between the points $(4,0)$ and $(-4,0)$ = 2
Hence $\sqrt{(x-4)^2+(y-0)^2}-\sqrt{(x+4)^2+(y-0)^2}=2$
$\Rightarrow \sqrt{(x-4)^2+y^2}=2+\sqrt{(x+4)^2+y^2}$
Squaring on both sides we get,
$(x-4)^2+y^2=4+(x+4)^2+y^2+4\sqrt{(x+4)^2+y^2}$
On expanding we get,
$x^2-8x+16+y^2=4+x^2+8x+16+y^2+4\sqrt{(x+4)^2+y^2}$
$-16x-4=4\sqrt{(x+4)^2+y^2}$
$-4(4x+1)=4\sqrt{(x+4)^2+y^2}$
$\Rightarrow -(4x+1)=\sqrt{(x+4)^2+y^2}$
Squaring on both sides we get,
$16x^2+8x+1=x^2+8x+16+y^2$
$15x^2-y^2=15$
Dividing throughout by 15 we get,
$\large\frac{x^2}{1}-\frac{y^2}{15}$$=1$
This is clearly the equation of a hyperbola.
Hence proved.
answered Oct 14, 2014 by sreemathi.v
 

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