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Pls solve this question as soon as possible of ellipse chapter

Pls solve this question as soon as possible of ellipse chapter
 

Question Image
Grade:12th pass

1 Answers

Aman Tomar
17 Points
one month ago
As the normals are perpendicular, therefore the corresponding tangents are also perpendicular. So they must intersect on the director circle which is given by x 
2
 +y 
2
 =a 
2
 +b 
2
 =A 
2
  (say)
Any point P on the director circle may be taken as, (Acosθ,Asinθ)
Let the equation to the ellipse be  
2
 
2
 
 + 
2
 
2
 
 =1
So the polar of P w.r.t. the ellipse will be  
2
 
Axcosθ
 + 
2
 
Aysinθ
 =1
The intersection of the perpendicular normals may be given by the equation
So putting l= 
2
 
Acosθ
  and m= 
2
 
Asinθ
  we get,
x=(a 
2
 −b 
2
 ) 
2
 
Acosθ
  
2
 
2
 +b 
2
 
 −cos 
2
 θ
1− 
2
 
(a 
2
 +b 
2
 )
 sin 
2
 θ
 +{ 
2
 
(a 
2
 +b 
2
 )sin 
2
 θ
 }
∴x= 
2
 +b 
2
 
2
 −b 
2
 
  
2
 cos 
2
 θ+a 
2
 sin 
2
 θ
2
 cos 
2
 θ−a 
2
 sin 
2
 θ
 Acosθ....1
and y=−(a 
2
 −b 
2
 ) 
2
 
Asinθ
  
2
 
(a 
2
 +b 
2
 )
 cos 
2
 θ+ 
2
 
(a 
2
 +b 
2
 )
 sin 
2
 θ
1−( 
2
 
2
 +b 
2
 
 )cos 
2
 θ
 
2
 +b 
2
 
2
 −b 
2
 
  
2
 cos 
2
 θ+a 
2
 sin 
2
 θ
2
 cos 
2
 θ−a 
2
 sin 
2
 θ
 Asinθ.....2
y
x
 = 
sinθ
cosθ
 
y
sinθ
 = 
x
cosθ
 = 
2
 +y 
2
 
 
1
 .....3
Eliminating from θ from 1 and 3, we getx= 
2
 +b 
2
 
2
 −b 
2
 
 A 
2
 +y 
2
 
 
x
  
2
 x 
2
 +a 
2
 y 
2
 
2
 x 
2
 −a 
2
 y 
2
 
 
or (a 
2
 +b 
2
 ) 
2
 (x 
2
 +y 
2
 )(b 
2
 x 
2
 +a 
2
 y 
2
 ) 
2
 =(a 
2
 −b 
2
 ) 
2
 (b 
2
 x 
2
 −a 
2
 y 
2
 ) 
2

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