Pls solve this question as soon as possible of ellipse chapter
Rohan Mittal , 3 Years ago
Grade 12th pass
Analytical Geometry> Pls solve this question as soon as possib...
1 Answers
Aman Tomar
Last Activity: 3 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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