# Pls solve this question as soon as possible of ellipse chapter

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
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