TS EAMCET 6-Aug-2021 Shift 1 Question Paper

Given equation of ellipse is ‌

x2

a2

+‌

y2

b2

=1 . . . (i)
and equation of parabola is y2=8ax ... (ii)
Differentiating Eq. (i) w.r.t x, we get, ‌

2x

a2

+‌

2yy′

b2

=0
⇒‌‌y′=−‌

b2x

a2y

Again, differentiating Eq. (ii) w.r.t x, we get,
y′=‌

4a

y

Since, curves intersect orthogonally,
‌

−b2x

a2y

×‌

4a

y

=−1⇒‌

4b2x

ay2

=1
⇒‌‌‌

4b2x

a(8ax)

=1
⇒‌‌‌

b2x

2a2x

=1⇒‌

a2

b2

=‌

1

2

∴‌‌e2=1−‌

a2

b2

=1−‌

1

2

=‌

1

2

∴‌‌(e2)2=(‌

1

2

)2=‌

1

4