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Question: A point moves such that the sum of its distances from two fixed points (ae,0) and (–ae,0) is always ...

A point moves such that the sum of its distances from two fixed points (ae,0) and (–ae,0) is always 2a. Then equation of its locus is.

A

x2a2+y2a2(1−e2)=1\frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2}(1 - e^{2})} = 1

B

x2a2−y2a2(1−e2)=1\frac{x^{2}}{a^{2}} - \frac{y^{2}}{a^{2}(1 - e^{2})} = 1

C

x2a2(1−e2)+y2a2=1\frac{x^{2}}{a^{2}(1 - e^{2})} + \frac{y^{2}}{a^{2}} = 1

D

None of these

Answer

x2a2+y2a2(1−e2)=1\frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2}(1 - e^{2})} = 1

Explanation

Solution

Let A(ae,0)A ( a e , 0 ) and B(−ae,0)B ( - a e , 0 ) be two given points and (h,k)( h , k ) be the coordinates of the moving point P.

Now PA+PB=2aP A + P B = 2 a

⇒(h−ae)2+k2+(h+ae)2+k2=2a\Rightarrow \sqrt { ( h - a e ) ^ { 2 } + k ^ { 2 } } + \sqrt { ( h + a e ) ^ { 2 } + k ^ { 2 } } = 2 a .....(i)

But, we know that

[(h−ae)2+k2]−[(h+ae)2+k2]=−4aeh\left[ ( h - a e ) ^ { 2 } + k ^ { 2 } \right] - \left[ ( h + a e ) ^ { 2 } + k ^ { 2 } \right] = - 4 a e h .....(ii)

Dividing (ii) by (i), we get

[(h−ae)2+k2]−[(h+ae)2+k2]=−2eh\sqrt { \left[ ( h - a e ) ^ { 2 } + k ^ { 2 } \right] } - \sqrt { \left[ ( h + a e ) ^ { 2 } + k ^ { 2 } \right] } = - 2 e h .....(iii)

Adding (i) and (iii),

2[h−ae)2+k2]=2(a−eh)2 \sqrt { \left. [ h - a e ) ^ { 2 } + k ^ { 2 } \right] } = 2 ( a - e h )

Squaring both sides, we get

⇒(h−ae)2+k2=(a−eh)2⇒h2a2+k2a2(1−e2)=1\Rightarrow ( h - a e ) ^ { 2 } + k ^ { 2 } = ( a - e h ) ^ { 2 } \Rightarrow \frac { h ^ { 2 } } { a ^ { 2 } } + \frac { k ^ { 2 } } { a ^ { 2 } \left( 1 - e ^ { 2 } \right) } = 1

Hence locus of P is x2a2+y2a2(1−e2)=1\frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { a ^ { 2 } \left( 1 - e ^ { 2 } \right) } = 1 .