TS EAMCET 4-Aug-2021 Shift 1 Question Paper

Equation of parabola
(y−k)2=4a(x−h). . . (i)
Where focus (h+a,k) and vertex (h,k)
Given, vertex (1,−2)=(h,k)
h‌‌=1,k=−2
⇒‌‌‌ Focus ‌(‌

5

4

,−2)‌‌=(h+a,k)
h+a‌‌=‌

5

4

a‌‌=‌

5

4

−1=‌

1

4

Putting values in Eq. (i), we get
‌(y+2)2=4⋅‌

1

4

(x−1)
‌(y+2)2=(x−1)
Clearly, point (10,1) satisfies this equation.
∴ Options (d) is correct.