If f1 and f2 be the feet of the perpendicular from foci s1 and s2 of an ellipse (x²/5) + (y²/3) =1 on the tangent at any point P on the ellipse, then (s1f1)(s2f2) is equal to ?

Ans --3

Any tangent to the ellipse has equation y = mx + a2m2+b2
And the foci have the coordinate as S1( ae , 0) and  S2 (-ae , 0)
Let the distance from S1 to the tangent be f1 and S2 to the tangent be f2
So S1f1 = aem +a2m2+b2m2+1
And S2f2 =-aem +a2m2+b2m2+1

So the product of (S1f1)*(S2f2) =
aem +a2m2+b2m2+1×-aem +a2m2+b2m2+1 =a2m2+b2 -a2e2m2m2+1 =a2m2(1-e2) +b2m2+1  =b2(m2+1)m2+1 [ as a2(1-e2) =b2]= b2

Hence b= 3  (ans)

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