Squaring on both sides, we get ‌(x−2)2+(y+3)2=‌
(2x+y−5)2
9
‌⇒9((x−2)2+(y+3)2) ‌=4x2+y2+25+4xy−20x−10y ‌=9x2−36x+36+9y2+54y+81 ‌=4x2+y2+25+4xy−20x−10y ‌⇒5x2+8y2−4xy−16x+64y+92=0 This is equation of ellipse. Now, slope of directrix is mD=−2 and the line joining the two foci is perpendicular to the directrix. So, mF=‌
−1
mD
=‌
−1
−2
=‌
1
2
Let, the other focus be F2(x2,y2) So, ‌
y2−(−3)
x2−2
=‌
y2+3
x2−2
=‌
1
2
‌⇒2(y2+3)=x2−2 ‌⇒‌‌x2=2y2+8 We know that c=ae and the distance between the focus and directrix is ‌‌
a
e
−c=‌
a
e
−ae=‌
a(1−e2)
e
‌⇒‌‌‌
a(1−(
√5
3
)2)
√5
3
=‌
4
√5
‌⇒‌‌‌
a(1−
5
9
)
√5
3
=‌
4
√5
⇒a(‌
4
9
)×‌
3
√5
=‌
4
√5
‌⇒‌‌‌
4a
3√5
=‌
4
√5
⇒a=3 ‌∴‌‌c=ae=3×‌
√5
3
=√5 The distance between two foci =2c=2√5 Let F2(x2,y2). So, the distance ‌F1F2=√(x2−2)2+(y2+3)2=2√5 ‌⇒√(2y2+8−2)2+(y2+3)2=2√5 ‌‌‌(∵x2=2y2+8) ‌⇒‌‌√4(y2+3)2+(y2+3)2=2√5 ‌⇒‌‌√5(y2+3)2=2√5 ‌⇒‌‌√5|y2+3|=2√5⇒|y2+3|=2 So, y2+3=2 or y2+3=−2 ⇒‌‌y2=−1‌ or ‌y2=−5 ∴‌x2=2(−1)+8=6‌ and ‌ ‌x2=2(−5)+8=−2 So, F2(6,−1) and F2(−2,−5) Line 2x+y−5=0 For F1(2,−3),2(2)+(−3)−5=4−8=−4 For F2(6,−1),2(6)+(−1)−5=12−6=6 Since, -4 and 6 have opposite signs, so foci of an ellipse is F2(−2,−5).