IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

rachit12

a point moves such that the sum of its distances from two fixed pts (ae,0) nd (-ae,0) is always 2a .prove that the equation of the locus is x²/a²+ y²/a²(1-e²) = 1..

cutepooja

k i will try

p(h;k)

A anB r (ae,0) nd (-ae,0)

PA+PB=2a

(h-ae)2+k2= 4a2 +(h+ae)2+k2 -4a root[(h+ae)2 +k2]

solving

(he +a)2 = (h + ae)2 + k2

contd

bt am i goin correct

cutepooja

e2h2 + a2+ 2hea = h2 +k2 + a2e2 + 2hea
h2[1-e2] + k2 = a2(1-e2)
replace h by x
and k by y
is it ok

computer001

the qsn stated by rachit is the defn of an ellipse...so nothing to prove in it

adieu

COMPUTER001 is correct

rachit12

(h-ae)2+k2= 4a2 +(h+ae)2+k2 -4a root[(h+ae)2 +k2]....

oye yeh kahan se aaya ??

cutepooja

WAIT I NEED TIME TO WRITE MY ANSWER

cutepooja

PA +PB = ROOTOF [ {h-ae}2+ k2 ] +ROOT [ {h + ae }2 +k2 ] = 2a

=>root (h-ae)2 + k2 = 2a - root[ (h+ ae)2 +k2]
squaring both sides
(h-ae)2 + k2 = 4a2 + k2 - 4aroot of[(h+ae)2 +k2] + (h+ae)2

=> - 4hea -4a2 + -4a root of [ (h+ae)2 +k2 ]
=> (he +a)2 = (h +ae)2 +k2
age bhi hai wait

cutepooja

=> e2h2 + a2+ 2hea = h2 +k2 + a2e2 + 2hea
h2[1-e2] + k2 = a2(1-e2)
=> h2/a2 + k2/a2(1-e2) = 1

rachit12

thanxx !!
;D..

ankur.kkhurana

bitsat was not in 1999

rachit12

arre bhai ml khanna main likha hain !!
jaake dekh le !!

heavensablaze

suppose the point is (x,y), then

$\sqrt{{(x-ae)}^{2}+{y}^{2}}+\sqrt{({x+ae)}^{2}+{y}^{2}}=2a$ (1)

Now

$[{(x-ae)}^{2}+{y}^{2}] - [({(x+ae)}^{2}+{y}^{2}] = 4aex$ (2)

On dividing (2) by(1)

$\sqrt{{(x-ae)}^{2}+{y}^{2}} - \sqrt{{(x+ae)}^{2}+{y}^{2}} = -2ex$ (3)

Adding (1) and (3)

$2\sqrt{{(x-ae)}^{2}+{y}^{2}} =2(a-ex)$

Squaring

x²-2aex+a²e² +y²=a²-2aex+e²x²

x²(1-e²)+y²=a(1-e²)

or

$\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{a}^{2}(1-{e}^{2})}$

Hope its useful

rachit12

arre bhai same is written in ml khanna !!
mujhe second eqn nahi samajh aayi thi !!??

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