TS EAMCET 6-Aug-2021 Shift 2 Question Paper

Let P be (at2,2at) and Q be (h,k).
Also given, A=(−2,3)
Now, since Q divides AP in the ratio 3:2 i.e.,
‌

AQ

QP

=‌

3

2

So, ‌‌h=‌

3at2−4

5

. . . (i)
and k=‌

6at+6

5

. . . (ii)
⇒‌‌5k=6at+6
⇒‌‌t=‌

5k−6

6a

put in Eq. (i), we get
5h=3a(‌

5k−6

6a

)2−4
⇒‌‌5h+4=‌

(5k−6)2

12a

⇒(5k−6)2=60ah+48a
⇒‌‌(5k−6)2=12a(5h+4)
Hence, the locus of Q(h,k) is
(5y−6)2=12a(5x+4) . . . (iii)
On comparing Eq. (iii) by Y2=4AX, we get,
⇒‌‌Y‌=5y−6,X=5x+4‌ and ‌4A=12a
A‌=3a
So, focus of Y2=4AX is (A,0) i.e. X=A and Y=0
Hence, focus of Eq. (iii) is
5x+4‌‌=3a‌ and ‌5y−6=0
x‌‌=‌

3a−4

5

‌ and ‌y=‌

6

5

i.e. ‌‌(x,y)=(‌

3a−4

5

,‌

6

5

)= Focus