AP EAMCET Engineering 23 Apr 2019 Shift 1 Paper

The equation of normal to hyperbola xy = 4 at point (2t,

2

t

) is,
2t4−xt3+yt−2=0 ....(I)
Above equation passes through (a, b ) so,
2t4−at3+bt−2=0
Have roots

α1

2

,

α2

2

,

α3

2

,

α4

2

so,

1

2

(α1+α2+α3+α4)=

a

2

α1+α2+α3+α4=a
∑(

α1α2

4

)=0
∑(

α1α2α3

8

)=

− b

2

and

α1α2α3α4

16

=−1
This implies,
β1+β2+β3+β4=

4

α1

+

4

α2

+

4

α3

+

4

α4

=4

α1α2α3

α1α2α3α4

=4

− b 2 ×8

−16

=b
This implies,

(α1+α2+α3+α4)

β1+β2+β3+β4

=(α1α2α3α4)
=

a(−16)

b

=−16

a

b