TS EAMCET 6-Aug-2021 Shift 1 Question Paper

‌

x2

18

+‌

y2

8

=1 . . . (i)
‌ and ‌x2−y2=5 . . . (ii)
∴‌‌x2=5+y2‌ in ellipse ‌
‌

5+y2

18

+‌

y2

8

=1⇒40+8y2+18y2=144
⇒‌‌26y2=104⇒y=±2
∴‌‌x2=5+y2=9⇒x=±3
Hence, A=(2,3),B=(−2,3),C=(−2,−3),
D=(2,−3)
Differentiating Eq. (i) w.r.t. x,⇒‌

x

9

+‌

y⋅y′

4

=0
⇒‌‌y′=−‌

4x

9y

∴‌‌(y′)(2,3)=−‌

8

27

(for ellipse)
Differentiating Eq. (ii) w.r.t. x,
2x−2yy′=0⇒y′=x∕y
=|‌

18−8

27

×‌

27×3

81+16

|=|‌

10×3

97

|
⇒‌‌tan‌θ1=30∕97
For ellipse, (y′)(−2,3)=‌

−4(−2)

9(3)

=‌

8

27

For hyperbola, (y′)(−2,3)=‌

3

−2

=−‌

2

3

∴tan‌θ2=|‌

‌ 8 27 − 2 3

1−‌ 8 27 ×(−‌ 2 3 )

|=|−‌

30

97

| (∴tan‌θ2= negative )
For ellipse, (y′)(−2,−3)=−‌

8

27

For hyperbola, (y′)(−2,−3)=2∕3
∴‌‌tan‌θ3=|‌

‌ 2 3 − 8 27

1−(‌ 2 3 )(−‌ 8 27 )

|=‌

30

97

For ellipse, (y′)(2−3)=8∕27
and for hyperbola, (y′)(2,−3)=‌

8

−3

∴‌‌tan‌θ4=|−‌

30

97

|=‌

30

97

Hence, θ1=θ2=θ3=θ4.