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Home » A point moves as so that the difference of its distances from (ae, 0) and (-ae, 0) is 2a, prove that the equation to its locus is
A point moves as so that the difference of its distances from (ae, 0) and (-ae, 0) is 2a, prove that the equation to its locus is
Brief review of Cartesian System of Rectangular Coordinates | CBSE | Class 11 | Exercise 22.2 | Maths | RD Sharma
A point moves as so that the difference of its distances from (ae, 0) and (-ae, 0) is 2a, prove that the equation to its locus is

RD Sharma Solutions for Class 11 Maths Chapter 22- image 14, where b2 = a2 (e2 – 1).

Solution:

Let P (h, k) represent any point on the locus

And let the coordinates of A and B be given by (ae, 0) and(-ae, 0).

Where, we have:

PA – PB = 2a

RD Sharma Solutions for Class 11 Maths Chapter 22- image 15

Upon squaring both sides we get:

{{\left( \mathbf{eh}\text{ }+\text{ }\mathbf{a} \right)}^{\mathbf{2}}}~=\text{ }{{\left( \mathbf{h}\text{ }+\text{ }\mathbf{ae} \right)}^{\mathbf{2}}}~+\text{ }{{\left( \mathbf{k}-\mathbf{0} \right)}^{\mathbf{2}}}

{{\mathbf{e}}^{\mathbf{2}}}{{\mathbf{h}}^{\mathbf{2}}}~+\text{ }{{\mathbf{a}}^{\mathbf{2}}}~+\text{ }\mathbf{2aeh}\text{ }=\text{ }{{\mathbf{h}}^{\mathbf{2}}}~+\text{ }{{\mathbf{a}}^{\mathbf{2}}}{{\mathbf{e}}^{\mathbf{2}}}~+\text{ }\mathbf{2aeh}\text{ }+\text{ }{{\mathbf{k}}^{\mathbf{2}}}{{\mathbf{h}}^{\mathbf{2}}}~

\left( {{\mathbf{e}}^{\mathbf{2}}}-\mathbf{1} \right)-{{\mathbf{k}}^{\mathbf{2}}}~=\text{ }{{\mathbf{a}}^{\mathbf{2}}}~({{\mathbf{e}}^{\mathbf{2}}}-\mathbf{1})

RD Sharma Solutions for Class 11 Maths Chapter 22- image 16

Upon replacing (h, k) with (x, y)

The locus of a point such that the difference of its distances from (ae, 0) and (-ae, 0) is 2a.

RD Sharma Solutions for Class 11 Maths Chapter 22- image 17
Where b2 = a2 (e2 – 1)

Hence proved.

i

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