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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
 

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Grade:12th pass

1 Answers

Aman Tomar
17 Points
2 years 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  
a 
2
 
x 
2
 
​
 + 
b 
2
 
y 
2
 
​
 =1
So the polar of P w.r.t. the ellipse will be  
a 
2
 
Axcosθ
​
 + 
b 
2
 
Aysinθ
​
 =1
The intersection of the perpendicular normals may be given by the equation
So putting l= 
a 
2
 
Acosθ
​
  and m= 
b 
2
 
Asinθ
​
  we get,
x=(a 
2
 −b 
2
 ) 
a 
2
 
Acosθ
​
  
a 
2
 
a 
2
 +b 
2
 
​
 −cos 
2
 θ
1− 
b 
2
 
(a 
2
 +b 
2
 )
​
 sin 
2
 θ
​
 +{ 
b 
2
 
(a 
2
 +b 
2
 )sin 
2
 θ
​
 }
∴x= 
a 
2
 +b 
2
 
a 
2
 −b 
2
 
​
  
b 
2
 cos 
2
 θ+a 
2
 sin 
2
 θ
b 
2
 cos 
2
 θ−a 
2
 sin 
2
 θ
​
 Acosθ....1
and y=−(a 
2
 −b 
2
 ) 
b 
2
 
Asinθ
​
  
a 
2
 
(a 
2
 +b 
2
 )
​
 cos 
2
 θ+ 
b 
2
 
(a 
2
 +b 
2
 )
​
 sin 
2
 θ
1−( 
a 
2
 
a 
2
 +b 
2
 
​
 )cos 
2
 θ
​
 
= 
a 
2
 +b 
2
 
a 
2
 −b 
2
 
​
  
b 
2
 cos 
2
 θ+a 
2
 sin 
2
 θ
b 
2
 cos 
2
 θ−a 
2
 sin 
2
 θ
​
 Asinθ.....2
y
x
​
 = 
sinθ
cosθ
​
 
y
sinθ
​
 = 
x
cosθ
​
 = 
x 
2
 +y 
2
 
​
 
1
​
 .....3
Eliminating from θ from 1 and 3, we getx= 
a 
2
 +b 
2
 
a 
2
 −b 
2
 
​
 A 
x 
2
 +y 
2
 
​
 
x
​
  
b 
2
 x 
2
 +a 
2
 y 
2
 
b 
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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