Let the given focus be
F1(2,−3) and the corresponding directrix be
L1:2x+y−5=0 and its eccentricity,
e=Let the other focus be
F2(x2,y2)Now, the distance from
F1(2,−3) to
2x+y−5=0 is
d1====Let
P(x,y) be any point on the ellipse.
So,
PF1=√(x−2)2+(y+3)2And the distance from
P to the directrix
L1 is
PM1=⇒Since,
PF1=e⋅PM1⇒√(x−2)2+(y+3)2=⋅⇒Squaring on both sides, we get
(x−2)2+(y+3)2=⇒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=0This 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===Let, the other focus be
F2(x2,y2)So,
== ⇒2(y2+3)=x2−2⇒x2=2y2+8We know that
c=ae and the distance between the focus and directrix is
−c=−ae=⇒=⇒=⇒a()×=⇒=⇒a=3∴c=ae=3×=√5The distance between two foci
=2c=2√5Let
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|=2So,
y2+3=2 or
y2+3=−2⇒y2=−1 or y2=−5∴x2=2(−1)+8=6 and x2=2(−5)+8=−2So,
F2(6,−1) and
F2(−2,−5)Line
2x+y−5=0For
F1(2,−3),2(2)+(−3)−5=4−8=−4For
F2(6,−1),2(6)+(−1)−5=12−6=6Since, -4 and 6 have opposite signs, so foci of an ellipse is
F2(−2,−5).
