Show that the product of the lengths of theperpendicular segments drawn from thefoci to any tangent line to the ellipsex + y2 = 1 is equal to 16.1625 About the author Ximena
Answer: + 16 y 2 =1e= 5 41 ae= 41 Foci S(ae,o)=S(41,o)S(ae,o)=S( 41 ,o) y=mx+ a 2 m 2 b 2 y=mx+ 25m 2 −16 P 1 = ∣ ∣ ∣ ∣ ∣ ∣ m 2 +1 m( 41 )+(−1)(o)+ 25m 2 −16 ∣ ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ ∣ m 2 +1 25m 2 −16 ∣ ∣ ∣ ∣ ∣ ∣ P 0 = ∣ ∣ ∣ ∣ ∣ ∣ m 2 +1 m(− 41 )+(−1)(o)+ 25m 2 − 16 ∣ ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ ∣ m 2 +1 25m 2 −16 − 41 m ∣ ∣ ∣ ∣ ∣ ∣ P 1 P 2 = ∣ ∣ ∣ ∣ ∣ ∣ m 2 +1 25m 2 −16 + 41 m ∣ ∣ ∣ ∣ ∣ ∣ × ∣ ∣ ∣ ∣ ∣ ∣ m 2 +1 25m 2 −16− 41 m ∣ ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ m 2 +1 25m 2 −16−41m 2 ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ m 2 +1 −16m 2 −16 ∣ ∣ ∣ ∣ ∣ =16 Reply
Answer:
+
16
y
2
=1e=
5
41
ae=
41
Foci S(ae,o)=S(41,o)S(ae,o)=S(
41
,o)
y=mx+
a
2
m
2
b
2
y=mx+
25m
2
−16
P
1
=
∣
∣
∣
∣
∣
∣
m
2
+1
m(
41
)+(−1)(o)+
25m
2
−16
∣
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∣
∣
m
2
+1
25m
2
−16
∣
∣
∣
∣
∣
∣
P
0
=
∣
∣
∣
∣
∣
∣
m
2
+1
m(−
41
)+(−1)(o)+
25m
2
−
16
∣
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∣
∣
m
2
+1
25m
2
−16
−
41
m
∣
∣
∣
∣
∣
∣
P
1
P
2
=
∣
∣
∣
∣
∣
∣
m
2
+1
25m
2
−16
+
41
m
∣
∣
∣
∣
∣
∣
×
∣
∣
∣
∣
∣
∣
m
2
+1
25m
2
−16−
41
m
∣
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∣
m
2
+1
25m
2
−16−41m
2
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∣
m
2
+1
−16m
2
−16
∣
∣
∣
∣
∣
=16