第12章解析幾何Analytic Geometry

Locating the Vertices and Foci of a Hyperbola双曲線の頂点と焦点の位置を求める

In analytic geometry, a hyperbola is a conic section formed by intersecting a right circular cone with a plane at an angle such that both halves of the cone are intersected. This intersection produces two separate unbounded curves that are mirror images of each other. See Figure 2.

解析幾何では、双曲線とは、直円錐を、円錐の両方の半分と交わるような角度の平面で切ってできる円錐曲線である。この交わりは、互いに鏡像である二本の別々の限りなく続く曲線を生む。図2を見よ。

An illustration showing a double cone (two cones joined at their vertices) being intersected by a vertical plane. The plane cuts through both parts of the double cone, creating two distinct, open curves (highlighted in orange) that together form a hyperbola. — Figure 2 A hyperbola図2 双曲線

Like the ellipse, the hyperbola can also be defined as a set of points in the coordinate plane. A hyperbola is the set of all points $(x, y)$ in a plane such that the difference of the distances between $(x, y)$ and the foci is a positive constant.

楕円と同じく、双曲線も座標平面の点の集合として定められる。双曲線とは、平面上の点 $(x, y)$ のうち、 $(x, y)$ と焦点との距離の差が正の定数であるものすべての集合である。

Notice that the definition of a hyperbola is very similar to that of an ellipse. The distinction is that the hyperbola is defined in terms of the difference of two distances, whereas the ellipse is defined in terms of the sum of two distances.

双曲線の定めが楕円の定めとたいへんよく似ていることに注意せよ。違いは、双曲線が二つの距離の差で定められるのに対し、楕円は二つの距離の和で定められる点である。

As with the ellipse, every hyperbola has two axes of symmetry. The transverse axis is a line segment that passes through the center of the hyperbola and has vertices as its endpoints. The foci lie on the line that contains the transverse axis. The conjugate axis is perpendicular to the transverse axis and has the co-vertices as its endpoints. The center of a hyperbola is the midpoint of both the transverse and conjugate axes, where they intersect. Every hyperbola also has two asymptotes that pass through its center. As a hyperbola recedes from the center, its branches approach these asymptotes. The central rectangle of the hyperbola is centered at the origin with sides that pass through each vertex and co-vertex; it is a useful tool for graphing the hyperbola and its asymptotes. To sketch the asymptotes of the hyperbola, simply sketch and extend the diagonals of the central rectangle. See Figure 3.

楕円と同じく、どの双曲線にも対称軸が二つある。主軸は双曲線の中心を通り、頂点を端点とする線分である。焦点は主軸を含む直線上にある。共役軸は主軸に垂直で、余頂点を端点とする。双曲線の中心は主軸と共役軸の両方の中点で、そこで両軸が交わる。どの双曲線にも、中心を通る二本の漸近線がある。双曲線が中心から遠ざかるにつれ、その枝はこの漸近線に近づく。双曲線の中心の長方形は原点を中心とし、各頂点と余頂点を通る辺を持つ。これは双曲線とその漸近線のグラフを描くのに役立つ道具である。双曲線の漸近線の概形を描くには、中心の長方形の対角線を描いて延ばすだけでよい。図3を見よ。

This diagram illustrates the various components of a hyperbola centered at the origin, including the vertices, co-vertices, foci, transverse axis, conjugate axis, asymptotes, and the rectangular box used to construct the asymptotes. — Figure 3 Key features of the hyperbola図3 双曲線の要となる特徴

In this section, we will limit our discussion to hyperbolas that are positioned vertically or horizontally in the coordinate plane; the axes will either lie on or be parallel to the x- and y-axes. We will consider two cases: those that are centered at the origin, and those that are centered at a point other than the origin.

この節では、座標平面に縦または横に置かれた双曲線に話を限る。軸は x軸と y軸の上にあるか、それらに平行である。二つの場合を考える。原点を中心とするものと、原点以外の点を中心とするものである。

Deriving the Equation of a Hyperbola Centered at the Origin原点を中心とする双曲線の式を導く

Let $(- c, 0)$ and $(c, 0)$ be the foci of a hyperbola centered at the origin. The hyperbola is the set of all points $(x, y)$ such that the difference of the distances from $(x, y)$ to the foci is constant. See Figure 4.

$(- c, 0)$ と $(c, 0)$ を、原点を中心とする双曲線の焦点とする。双曲線とは、 $(x, y)$ から焦点までの距離の差が一定である点 $(x, y)$ すべての集合である。図4を見よ。

A horizontal hyperbola in the x y coordinate system centered at (0, 0) with Vertices at (negative a, 0) and (a, 0) and Foci at (negative c, 0) and (c, 0), with lines of length d1 and d2 connecting a point on the right branch of the hyperbola to the foci.

If $(a, 0)$ is a vertex of the hyperbola, the distance from $(- c, 0)$ to $(a, 0)$ is $a - (- c) = a + c .$ The distance from $(c, 0)$ to $(a, 0)$ is $c - a .$ The difference of the distances from the foci to the vertex is

$(a, 0)$ がこの双曲線の頂点なら、 $(- c, 0)$ から $(a, 0)$ までの距離は $a - (- c) = a + c$ である。 $(c, 0)$ から $(a, 0)$ までの距離は $c - a$ である。焦点から頂点までの距離の差は次である。

(a + c) - (c - a) = 2 a

If $(x, y)$ is a point on the hyperbola, we can define the following variables:

$(x, y)$ が双曲線上の点なら、次の変数を定められる。

\begin{array}{l} d_{2} = the distance from (- c, 0) to (x, y) \\ d_{1} = the distance from (c, 0) to (x, y) \end{array}

By definition of a hyperbola, $d_{2} - d_{1}$ is constant for any point $(x, y)$ on the hyperbola. We know that the difference of these distances is $2 a$ for the vertex $(a, 0) .$ It follows that $d_{2} - d_{1} = 2 a$ 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.

双曲線の定めにより、 $d_{2} - d_{1}$ は双曲線上のどの点 $(x, y)$ についても一定である。頂点 $(a, 0)$ については、この距離の差が $2 a$ だと分かっている。よって双曲線上のどの点についても $d_{2} - d_{1} = 2 a$ である。楕円の式の導出と同じく、まず距離の公式を使う。あとの導出は代数である。この導出を、前の節の楕円のものと比べよ。

\begin{array}{l} d_{2} - d_{1} = \sqrt{{(x - (- c))}^{2} + {(y - 0)}^{2}} - \sqrt{{(x - c)}^{2} + {(y - 0)}^{2}} = 2 a & Distance Formula \\ \sqrt{{(x + c)}^{2} + y^{2}} - \sqrt{{(x - c)}^{2} + y^{2}} = 2 a & Simplify expressions . \\ \sqrt{{(x + c)}^{2} + y^{2}} = 2 a + \sqrt{{(x - c)}^{2} + y^{2}} & Move radical to opposite side . \\ {(x + c)}^{2} + y^{2} = {(2 a + \sqrt{{(x - c)}^{2} + y^{2}})}^{2} & Square both sides . \\ x^{2} + 2 c x + c^{2} + y^{2} = 4 a^{2} + 4 a \sqrt{{(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 \sqrt{{(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 \sqrt{{(x - c)}^{2} + y^{2}} - 2 c x & Combine like terms . \\ 4 c x - 4 a^{2} = 4 a \sqrt{{(x - c)}^{2} + y^{2}} & Isolate the radical . \\ c x - a^{2} = a \sqrt{{(x - c)}^{2} + y^{2}} & Divide by 4 . \\ {(c x - a^{2})}^{2} = a^{2} {(\sqrt{{(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} . \\ \frac{x^{2} b^{2}}{a^{2} b^{2}} - \frac{a^{2} y^{2}}{a^{2} b^{2}} = \frac{a^{2} b^{2}}{a^{2} b^{2}} & Divide both sides by a^{2} b^{2} \\ \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \end{array}

This equation defines a hyperbola centered at the origin with vertices $(\pm a, 0)$ and co-vertices $(0 \pm b) .$

この式は、原点を中心とし、頂点 $(\pm a, 0)$ 、余頂点 $(0 \pm b)$ を持つ双曲線を定める。

Standard Forms of the Equation of a Hyperbola with Center (0,0) 原点を中心とする双曲線の標準形

The standard form of the equation of a hyperbola with center $(0, 0)$ and transverse axis on the x-axis is

中心 $(0, 0)$ 、主軸が x軸上にある双曲線の式の標準の形は次である。

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

where

ここで

• the length of the transverse axis is $2 a$ • the coordinates of the vertices are $(\pm a, 0)$ • the length of the conjugate axis is $2 b$ • the coordinates of the co-vertices are $(0, \pm b)$ • the distance between the foci is $2 c,$ where $c^{2} = a^{2} + b^{2}$ • the coordinates of the foci are $(\pm c, 0)$ • the equations of the asymptotes are $y = \pm \frac{b}{a} x$

・主軸の長さは $2 a$ ・頂点の座標は $(\pm a, 0)$ ・共役軸の長さは $2 b$ ・余頂点の座標は $(0, \pm b)$ ・焦点どうしの距離は $2 c$ 。ここで $c^{2} = a^{2} + b^{2}$ ・焦点の座標は $(\pm c, 0)$ ・漸近線の式は $y = \pm \frac{b}{a} x$

See Figure 5a.

図5aを見よ。

The standard form of the equation of a hyperbola with center $(0, 0)$ and transverse axis on the y-axis is

中心 $(0, 0)$ 、主軸が y軸上にある双曲線の式の標準の形は次である。

\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1

where

ここで

• the length of the transverse axis is $2 a$ • the coordinates of the vertices are $(0, \pm a)$ • the length of the conjugate axis is $2 b$ • the coordinates of the co-vertices are $(\pm b, 0)$ • the distance between the foci is $2 c,$ where $c^{2} = a^{2} + b^{2}$ • the coordinates of the foci are $(0, \pm c)$ • the equations of the asymptotes are $y = \pm \frac{a}{b} x$

・主軸の長さは $2 a$ ・頂点の座標は $(0, \pm a)$ ・共役軸の長さは $2 b$ ・余頂点の座標は $(\pm b, 0)$ ・焦点どうしの距離は $2 c$ 。ここで $c^{2} = a^{2} + b^{2}$ ・焦点の座標は $(0, \pm c)$ ・漸近線の式は $y = \pm \frac{a}{b} x$

See Figure 5b.

図5bを見よ。

Note that the vertices, co-vertices, and foci are related by the equation $c^{2} = a^{2} + b^{2} .$ When we are given the equation of a hyperbola, we can use this relationship to identify its vertices and foci.

頂点、余頂点、焦点は式 $c^{2} = a^{2} + b^{2}$ で結ばれていることに注意せよ。双曲線の式が与えられたとき、この関係を使ってその頂点と焦点が見分けられる。

The left graph displays a hyperbola centered at the origin with a horizontal transverse axis. Its vertices are at (±a, 0), foci at (±c, 0), and its asymptotes are given by the equations y = ±(b/a)x. An auxiliary dashed rectangle, defined by x = ±a and y = ±b, is shown, with the asymptotes passing through its corners. The right graph illustrates a hyperbola centered at the origin with a vertical transverse axis. Its vertices are at (0, ±a), foci at (0, ±c), and its asymptotes are given by the equations y = ±(a/b)x. An auxiliary dashed rectangle, defined by x = ±b and y = ±a, is also shown, with the asymptotes passing through its corners. — Figure 5 (a) Horizontal hyperbola with center $(0, 0)$ (b) Vertical hyperbola with center $(0, 0)$ 図5 (a) 中心 $(0, 0)$ の横向きの双曲線 (b) 中心 $(0, 0)$ の縦向きの双曲線

How To 手順

Given the equation of a hyperbola in standard form, locate its vertices and foci.

双曲線の式が標準の形で与えられたとき、その頂点と焦点の位置を求める。

1. Determine whether the transverse axis lies on the x- or y-axis. Notice that $a^{2}$ is always under the variable with the positive coefficient. So, if you set the other variable equal to zero, you can easily find the intercepts. In the case where the hyperbola is centered at the origin, the intercepts coincide with the vertices. If the equation has the form $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1,$ then the transverse axis lies on the x-axis. The vertices are located at $(\pm a, 0),$ and the foci are located at $(\pm c, 0) .$ If the equation has the form $\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1,$ then the transverse axis lies on the y-axis. The vertices are located at $(0, \pm a),$ and the foci are located at $(0, \pm c) .$ 2. Solve for $a$ using the equation $a = \sqrt{a^{2}} .$ 3. Solve for $c$ using the equation $c = \sqrt{a^{2} + b^{2}} .$

1. 主軸が x軸上にあるか y軸上にあるかを決める。 $a^{2}$ はつねに係数が正の変数の下にあることに注意せよ。よって、もう一方の変数を零と置けば、座標軸との交点が容易に求まる。双曲線が原点を中心とする場合、この交点は頂点と一致する。式が $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ の形なら、主軸は x軸上にある。頂点は $(\pm a, 0)$ 、焦点は $(\pm c, 0)$ にある。式が $\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1$ の形なら、主軸は y軸上にある。頂点は $(0, \pm a)$ 、焦点は $(0, \pm c)$ にある。2. 式 $a = \sqrt{a^{2}}$ を使って $a$ について解く。3. 式 $c = \sqrt{a^{2} + b^{2}}$ を使って $c$ について解く。

Example 1例1

Locating a Hyperbola’s Vertices and Foci双曲線の頂点と焦点の位置を求める

Identify the vertices and foci of the hyperbola with equation $\frac{y^{2}}{49} - \frac{x^{2}}{32} = 1.$

式 $\frac{y^{2}}{49} - \frac{x^{2}}{32} = 1.$ の双曲線の頂点と焦点を見分けよ。

Solution 解答

The equation has the form $\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1,$ so the transverse axis lies on the y-axis. The hyperbola is centered at the origin, so the vertices serve as the y-intercepts of the graph. To find the vertices, set $x = 0,$ and solve for $y .$

この式は $\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1$ の形なので、主軸は y軸上にある。この双曲線は原点を中心とするので、頂点はグラフの y軸との交点になる。頂点を求めるには、 $x = 0$ と置いて $y$ について解く。

\begin{array}{l} 1 = \frac{y^{2}}{49} - \frac{x^{2}}{32} \\ 1 = \frac{y^{2}}{49} - \frac{0^{2}}{32} \\ 1 = \frac{y^{2}}{49} \\ y^{2} = 49 \\ y = \pm \sqrt{49} = \pm 7 \end{array}

The foci are located at $(0, \pm c) .$ Solving for $c,$

焦点は $(0, \pm c)$ にある。 $c$ について解くと、

c = \sqrt{a^{2} + b^{2}} = \sqrt{49 + 32} = \sqrt{81} = 9

Therefore, the vertices are located at $(0, \pm 7),$ and the foci are located at $(0, \pm 9) .$

したがって、頂点は $(0, \pm 7)$ 、焦点は $(0, \pm 9)$ にある。

Try It #1やってみよう1

Identify the vertices and foci of the hyperbola with equation $\frac{x^{2}}{9} - \frac{y^{2}}{25} = 1.$

式 $\frac{x^{2}}{9} - \frac{y^{2}}{25} = 1.$ の双曲線の頂点と焦点を見分けよ。