TS EAMCET 4-Aug-2021 Shift 1 Question Paper

Equation of ellipse
‌

(x−α)2

a2

+‌

(y−β)2

b2

=1
Where foci (α±ae,β)
Equation of directrix x=α±‌

a

e

, where e is eccentricity.
=(0,0)
‌ Given, ‌(α+ae,β)=(0,0)‌‌ [focus]‌
⇒‌α+ae=0,β=0
⇒‌α+a⋅‌

1

2

=0
‌ Given, ‌(α+ae,β)=(0,0)‌‌ [focus] ‌
⇒‌α+ae=0,β=0‌
⇒‌α+a⋅‌

1

2

=0‌...(i)‌[e=‌

1

2

]
‌ Also, ‌‌x=α+‌

a

e

=4‌‌ [directrix] ‌
⇒‌α+2a=4‌...‌ (ii) ‌
Solving Eqs. (i) and (ii), we get
a=‌

8

3

From Eq. (i), α=−‌

4

3

Also, ‌‌e2=1−‌

b2

a2

⇒‌‌

1

4

=1−‌

b2

64∕9

⇒‌b2=‌

16

3

∴ Equation of ellipse
‌

(x+ 4 3 )2

64 9

+‌

y2

16 3

=1
⇒‌‌(3x+4)2+12y2=64