Question

Tangents are drawn to ellipse x²/a²+y²/b²=1 ,(a>b) and circle x²+y²=a² at point where common ordinate cuts them (on same side of x-axis),Then greatest acute angle between
these tangents is given by ????

Answer 1

Hey there !!!!!

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Let tangent to ellipse x²/a²+y²/b²=1 touch the ellipse at a point on  ellipse "P"

Co-ordinates of P= (acosα,bsinα)  

Equation of tangent to ellipse is S₁=0

       xx₁/a²+yy₁/b²=1

x₁=acosα      y₁=bsinα

So, equation of ellipse is 

  xcosα/a+ysinα/b=1-----Equation 1

Let Q(acosα,asinα) be a point on circle x²+y²=a² ."Q" is the point where tangent touches the circle  x²+y²=a²

So, equation of tangent = S₁=0

    xx₁+yy₁=a²

    axcosα+aysinα=a²

    xcosα+ysinα=a-----Equation 2

  xcosα/a+ysinα/b=1 is of the form Ax+By=c

Slope of Ax+By=c is -A/B

Comparing  xcosα/a+ysinα/b=1 with Ax+By=c

A=cosα/a   B=sinα/b  

Slope = -cosα*b/sinα*a=-b*cotα/a

Let m₁= -bcotα/a

Similarly comparing     xcosα+ysinα=a with Ax+By=c

A=cosα   B=sinα

Slope= -cosα/sinα = -cotα

Let m₂= - cotα

Now these tangents to circle and ellipse  make an an angle ∅  with slopes 
m₁ and m₂

Tan∅= m₁-m₂/1+m₁m₂

m₁= -bcotα/a    m₂= - cotα

tan∅= -bcotα/a-(- cotα)/1+(-bcotα*- cotα/a)

tan∅=  cotα(1-b/a)/1+bcot²α/a

Further simplification gives

tan∅=(a-b)/a*tanα+b*cotα----Equation 3

Now a*tanα+b*cotα  can be written in form  A²+B²=(A-B)²+2AB

Here A= √a*tanα   B=√b*cotα

a*tanα+b*cotα=(√a*tanα-√b*cotα)²+2√a*tanα*√b*cotα

So equation 3 changes to 

tan∅= a+b/(√a*tanα-√b*cotα)²+2(√a*tanα*√b*cotα)

According to question greatest acute angle "∅" between tangents is possible only when 

 ∅=tan⁻¹(a+b/(√a*tanα-√b*cotα)²+2(√a*tanα*√b*cotα) is maximum

And it is maximum when denominator is minimum

(√a*tanα-√b*cotα)²+2(√a*tanα*√b*cotα) can be minimum when 

(√a*tanα-√b*cotα)²=0 because 2(√a*tanα*√b*cotα) is always positive

So (√a*tanα-√b*cotα)²=0

   √a*tanα=√b*cotα

    So, maximum value of acute angle ∅ :

    ∅=tan⁻¹(a+b/2√a*tanα*√b*cotα)

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Hope this helped you.............