state and prove the standard form of hyperbola
Answers
Answer:
A General Note: Standard Forms of the Equation of a Hyperbola with Center (0,0) Note that the vertices, co-vertices, and foci are related by the equation c 2 = a 2 + b 2 \displaystyle {c}^{2}={a}^{2}+{b}^{2} c2=a2+b2.
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Answer:Deriving the Equation of a Hyperbola Centered at the Origin
Let \displaystyle \left(-c,0\right)(−c,0) and \displaystyle \left(c,0\right)(c,0) be the foci of a hyperbola centered at the origin. The hyperbola is the set of all points \displaystyle \left(x,y\right)(x,y) such that the difference of the distances from \displaystyle \left(x,y\right)(x,y) to the foci
If \displaystyle \left(a,0\right)(a,0) is a vertex of the hyperbola, the distance from \displaystyle \left(-c,0\right)(−c,0) to \displaystyle \left(a,0\right)(a,0) is \displaystyle a-\left(-c\right)=a+ca−(−c)=a+c. The distance from \displaystyle \left(c,0\right)(c,0) to \displaystyle \left(a,0\right)(a,0) is \displaystyle c-ac−a. The sum of the distances from the foci to the vertex is
\displaystyle \left(a+c\right)-\left(c-a\right)=2a(a+c)−(c−a)=2a
If \displaystyle \left(x,y\right)(x,y) is a point on the hyperbola, we can define the following variables:
\displaystyle \begin{array}{l}{d}_{2}=\text{the distance from }\left(-c,0\right)\text{ to }\left(x,y\right)\\ {d}_{1}=\text{the distance from }\left(c,0\right)\text{ to }\left(x,y\right)\end{array}
d
2
=the distance from (−c,0) to (x,y)
d
1
=the distance from (c,0) to (x,y)
By definition of a hyperbola, \displaystyle {d}_{2}-{d}_{1}d
2
−d
1
is constant for any point \displaystyle \left(x,y\right)(x,y) on the hyperbola. We know that the difference of these distances is \displaystyle 2a2a for the vertex \displaystyle \left(a,0\right)(a,0). It follows that \displaystyle {d}_{2}-{d}_{1}=2ad
2
−d
1
=2a for any point on the hyperbola. As with the derivation of the equation of an ellipse, we will begin by applying the distance formula. The rest of the derivation is algebraic. Compare this derivation with the one from the previous section for ellipses.
d
2
−
d
1
=
√
(
x
−
(
−
c
)
)
2
+
(
y
−
0
)
2
−
√
(
x
−
c
)
2
+
(
y
−
0
)
2
=
2
a
Distance Formula
√
(
x
+
c
)
2
+
y
2
−
√
(
x
−
c
)
2
+
y
2
=
2
a
Simplify expressions
.
√
(
x
+
c
)
2
+
y
2
=
2
a
+
√
(
x
−
c
)
2
+
y
2
Move radical to opposite side
.
(
x
+
c
)
2
+
y
2
=
(
2
a
+
√
(
x
−
c
)
2
+
y
2
)
2
Square both sides
.
x
2
+
2
c
x
+
c
2
+
y
2
=
4
a
2
+
4
a
√
(
x
−
c
)
2
+
y
2
+
(
x
−
c
)
2
+
y
2
Expand the squares
.
x
2
+
2
c
x
+
c
2
+
y
2
=
4
a
2
+
4
a
√
(
x
−
c
)
2
+
y
2
+
x
2
−
2
c
x
+
c
2
+
y
2
Expand remaining square
.
2
c
x
=
4
a
2
+
4
a
√
(
x
−
c
)
2
+
y
2
−
2
c
x
Combine like terms
.
4
c
x
−
4
a
2
=
4
a
√
(
x
−
c
)
2
+
y
2
Isolate the radical
.
c
x
−
a
2
=
a
√
(
x
−
c
)
2
+
y
2
Divide by 4
.
(
c
x
−
a
2
)
2
=
a
2
[
√
(
x
−
c
)
2
+
y
2
]
2
Square both sides
.
c
2
x
2
−
2
a
2
c
x
+
a
4
=
a
2
(
x
2
−
2
c
x
+
c
2
+
y
2
)
Expand the squares
.
c
2
x
2
−
2
a
2
c
x
+
a
4
=
a
2
x
2
−
2
a
2
c
x
+
a
2
c
2
+
a
2
y
2
Distribute
a
2
.
a
4
+
c
2
x
2
=
a
2
x
2
+
a
2
c
2
+
a
2
y
2
Combine like terms
.
c
2
x
2
−
a
2
x
2
−
a
2
y
2
=
a
2
c
2
−
a
4
Rearrange terms
.
x
2
(
c
2
−
a
2
)
−
a
2
y
2
=
a
2
(
c
2
−
a
2
)
Factor common terms
.
x
2
b
2
−
a
2
y
2
=
a
2
b
2
Set
b
2
=
c
2
−
a
2
.
x
2
b
2
a
2
b
2
−
a
2
y
2
a
2
b
2
=
a
2
b
2
a
2
b
2
Divide both sides by
a
2
b
2
x
2
a
2
−
y
2
b
2
=
1
This equation defines a hyperbola centered at the origin with vertices \displaystyle \left(\pm a,0\right)(±a,0) and co-vertices \displaystyle \left(0\pm b\right)(0±b).
Step-by-step explanation: