TS EAMCET 9-Sep-2020 Shift 2 Solved Paper

We, have, P(1,k) lies on ellipse
9x2+4y2−18x+16y−11=0
∴‌‌9+4k2−18+16k−11=0
⇒‌‌4k2+16k−20=0
⇒‌‌k2+4k−5=0
⇒‌‌(k+5)(k−1)=0⇒k=−5,1
∴P is either (1,−5) or (1,1)
Again, equation of ellipse is
9x2+4y2−18x+16y−11=0
⇒‌‌9(x−1)2+4(y+2)2=36
⇒‌‌‌

(x−1)2

4

+‌

(y+2)2

9

=0
∴ Eccentricity, ∴ Equation of directrices are
y+2=±‌

b

e

⇒y+2=±‌

9

√5

⇒‌‌y+2+‌

9

√5

=0 or y+2−‌

9

√5

=0
Now, distance of (1,−5) from y+2+‌

9

√5

=0 is
|−5+2+‌

9

√5

|=‌

9

√5

−3
Distance of (1,−5) from y+2−‌

9

√5

=0 is
|−5+2−‌

9

√5

|=‌

9

√5

+3
Distance of (1,1) from y+2+‌

9

√5

=0 is |1+1+‌

9

√5

| =‌

9

√5

+2
and distance of (1,1) from y+2−‌

9

√5

=0
|1+2−‌

9

2√5

|=‌

9

√5

−3
∴ Shortest distance =‌

9

√5

−3