Hyperbola Geometry

01 / UNDERSTAND & APPLY

1. A fixed absolute distance difference produces two branches

A hyperbola is the locus of points for which the absolute difference of distances to two fixed foci is a positive constant. With foci (−c,0),(c,0), write the difference as 2a, where 0<a<c. The strict inequality is required for a nondegenerate hyperbola because the difference of distances cannot exceed the separation 2c.

The horizontal standard form is x²/a²−y²/b²=1 with a,b>0 and c²=a²+b². The vertices are (±a,0), the foci are (±c,0), and the center is the midpoint of either pair. The domain in x is x≤−a or x≥a, producing the left and right branches.

Unlike an ellipse, a is not defined to be the larger of a and b. It is the semi-transverse axis associated with the positive term. The number b determines the conjugate-axis scale and asymptote slopes, but the points (0,±b) are not points on this horizontal hyperbola.

For a vertical hyperbola, y²/a²−x²/b²=1, the positive term is y² and the vertices and foci lie on the y-axis. Normalize the right side to 1 before identifying the orientation; multiplying the whole equation by −1 can swap which term appears positive.

02 / UNDERSTAND & APPLY

2. Squaring must retain the chosen branch information

On the right branch, let d_L be the distance to the left focus and d_R the distance to the right focus. Then d_L−d_R=2a. Also d_L²−d_R²=4cx, so d_L+d_R=2cx/a. Solving gives d_L=(c/a)x+a and d_R=(c/a)x−a.

Substituting d_R into (x−c)²+y²=d_R² and simplifying yields x²/a²−y²/(c²−a²)=1. Set b²=c²−a²>0. On the right branch x≥a, the expression (c/a)x−a is at least c−a>0, so the distance sign is consistent.

The squared equation also contains the left branch. There the absolute distance difference is still 2a, but its signed order reverses: d_R−d_L=2a. Thus an original condition with a signed difference selects one branch, whereas an absolute difference selects both.

The limiting cases require separate interpretation. If the constant difference is zero, the locus is the perpendicular bisector of the foci rather than a nondegenerate hyperbola. If it equals the focal separation, the triangle inequality becomes equality and the locus degenerates into outer rays on the focal axis.

03 / UNDERSTAND & APPLY

3. The asymptotes come from the leading quadratic directions

For x²/a²−y²/b²=1, the asymptotes are y=±(b/a)x. The homogeneous equation x²/a²−y²/b²=0 gives these two lines, but the hyperbola itself has right side 1. Therefore no finite point lies on both the standard hyperbola and either asymptote.

To verify the asymptotic approach on the upper right branch, write y=(b/a)√(x²−a²). The difference from (b/a)x is −ab/[√(x²−a²)+x], which tends to zero as x→+∞. Other tails follow by symmetry and the correct signs.

For center (h,k), replace x,y by x−h,y−k. The horizontal asymptotes become y−k=±(b/a)(x−h), passing through the center. For a vertical hyperbola the slopes are ±a/b instead.

The auxiliary rectangle with corners (±a,±b) in centered coordinates helps draw the asymptotes as its diagonals. It does not represent a bounded region enclosed by the hyperbola, and the conjugate-axis endpoints are plotting guides rather than actual intercepts.

04 / UNDERSTAND & APPLY

4. Hyperbolic eccentricity exceeds one

The eccentricity is e=c/a=√(1+b²/a²)>1. The directrices of the horizontal hyperbola are x=±a/e=±a²/c. Unlike ellipse directrices, these lie between the center and the vertices because a/e<a.

For the right focus (c,0) and right directrix x=a/e, the right branch satisfies PF=e(x−a/e)=ex−a. On the left branch x≤−a, the distance to this directrix is a/e−x and the same focus distance is a−ex. The absolute-distance formulation therefore describes both branches consistently.

The latus rectum through the right focus is the vertical chord x=c. Substitution gives y=±b²/a, so its length is 2b²/a, the same algebraic expression as for an ellipse even though the relation among a,b,c has a different sign.

The vertex-to-near-focus distance is c−a, positive because the focus lies farther from the center than the vertex. This is a quick geometry check: placing a hyperbola focus between the center and a vertex confuses it with the ellipse case.

05 / UNDERSTAND & APPLY

5. Parameters must cover the intended branch

The trigonometric parametrization x=a sec θ,y=b tan θ satisfies the equation through sec²θ−tan²θ=1. For −π/2<θ<π/2 it traces the right branch because sec θ>0. An interval with negative cosine traces the left branch. Values with cos θ=0 are excluded.

In normalized coordinates X=x/a, Y=y/b, intersect the line Y=t(X+1) through the known vertex (−1,0) with X²−Y²=1. One convenient form is x=a(1+t²)/(1−t²), y=2bt/(1−t²), with t≠±1. It satisfies the equation by the identity (1+t²)²−4t²=(1−t²)².

For |t|<1 this rational form lies on the right branch; for |t|>1 it lies on the left branch. The left vertex (−a,0) is omitted for finite t and can be viewed as the limiting point as |t|→∞. A parametrization need not cover every point with a finite parameter.

If hyperbolic functions are available, x=a cosh u,y=b sinh u covers the right branch for real u, with x=−a cosh u for the left. This is optional notation, not a requirement for the trigonometric and algebraic methods used here.

06 / UNDERSTAND & APPLY

6. The gradient supplies both tangent and normal

At a point P=(x₁,y₁) on x²/a²−y²/b²=1, the tangent is xx₁/a²−yy₁/b²=1. The normal direction is (x₁/a²,−y₁/b²), which is nonzero on the curve. These formulas follow by differentiating the defining equation or using its gradient.

At the vertices (±a,0), the tangent is vertical and the normal is horizontal. A formula obtained by solving for dy/dx divides by y and fails at those points, so retain the implicit line form for a complete description.

When x₁≠0, the normal slope is −a²y₁/(b²x₁). A center-to-point line generally has a different slope. As with an ellipse, circle intuition that every normal passes through the center is not valid.

The point must satisfy the hyperbola equation before the tangent-at-contact formula is used. Applying its algebraic expression to an arbitrary off-curve point produces a polar line rather than a tangent at that point.

07 / UNDERSTAND & APPLY

7. A zero discriminant test must still describe a quadratic

Substituting y=mx+d into the hyperbola produces a quadratic in x unless the line is parallel to an asymptote. For a genuine finite-slope tangent, the repeated-root calculation gives d²=a²m²−b², with |m|>b/a so that d is real and nonzero.

At |m|=b/a and d=0, the same formal equation d²=a²m²−b² holds, but the line is an asymptote and has no finite contact. The quadratic coefficient has vanished, so the repeated-root interpretation is no longer applicable. This exceptional case must be excluded.

A line parallel to an asymptote but with d≠0 can meet the hyperbola at exactly one finite point without being tangent. The second intersection is lost from the ordinary affine picture because the substituted equation becomes linear. Thus “exactly one intersection” alone is not a sufficient tangency test for an unbounded conic.

Vertical tangents x=±a are not represented by finite m and must be included separately. There are no horizontal tangents to a horizontal-transverse hyperbola because their proposed tangent intercept square would be negative.

08 / UNDERSTAND & APPLY

8. Perpendicular asymptotes define the rectangular case

A hyperbola is rectangular when its asymptotes are perpendicular. For x²/a²−y²/b²=1, their slopes have product −b²/a², so perpendicularity requires a=b. The eccentricity is then √2.

Rotating x²−y²=a² by 45 degrees converts it to a constant-product form uv=a²/2, with signs depending on the chosen rotation orientation. The common form xy=k, k≠0, has the coordinate axes as perpendicular asymptotes and center at the origin.

For xy=k, a point can be parametrized as (t,k/t), t≠0. Implicit differentiation gives y′=−y/x, and the tangent at (x₁,y₁) is y₁x+x₁y=2k. Its axis intercepts are 2x₁ and 2y₁, so the contact point is their midpoint.

The conjugate hyperbola of x²/a²−y²/b²=1 is y²/b²−x²/a²=1. It shares the same asymptotes but opens along the other axis. It is not obtained by merely exchanging the labels of the two branches of the original curve.

09 / UNDERSTAND & APPLY

9. Midpoint formulas require an actual real chord

For a chord with midpoint M=(x₀,y₀), substituting M+td into the quadratic shows that the linear coefficient must vanish: x₀dₓ/a²−y₀dᵧ/b²=0. Thus a noncentral candidate midpoint gives the line xx₀/a²−yy₀/b²=x₀²/a²−y₀²/b².

Unlike the ellipse case, not every point in a simple bounded interior region corresponds to a real chord. The resulting line must still be checked for two real intersections. The quadratic coefficient along the direction and its discriminant determine whether endpoints exist.

At the center, the midpoint equation becomes 0=0, so no direction is selected. Some lines through the center meet opposite branches in symmetric real points, while directions between the asymptotes may miss the curve. For a line y=mx through the origin, intersections exist precisely when |m|<b/a in the horizontal standard form.

Parallel chord midpoint loci again pass through the center, but exceptional directions parallel to asymptotes need separate treatment because their intersections are not ordinary two-endpoint chords. Keep existence conditions attached to algebraic midpoint relations.

10 / UNDERSTAND & APPLY

10. Factoring the quadratic exposes a constant product

The standard equation factors as (x/a+y/b)(x/a−y/b)=1. Define U=x/a+y/b and V=x/a−y/b. Then UV=1, while the inverse transformation is x=a(U+V)/2 and y=b(U−V)/2. This is an invertible linear coordinate change, so it loses no curve points.

The lines U=0 and V=0 are precisely the two asymptotes. Their product cannot equal 1 on either line, explaining directly why the curve never reaches an asymptote at a finite point. Far along a branch, one coordinate can become large while the other approaches zero.

Setting U=t and V=1/t gives x=a(t+1/t)/2 and y=b(t−1/t)/2 for t≠0. Positive t covers the right branch and negative t covers the left branch. Unlike the earlier rational parameter with denominator 1−t², this alternative includes both vertices at finite parameters t=1 and t=−1.

The transformed U,V coordinates are not generally an orthonormal Euclidean coordinate system. Their constant-product form is excellent for algebra, but lengths and angles must still be computed in the original x,y coordinates or with the transformed metric. Only in the rectangular case does a suitably normalized rotation align the asymptotes with perpendicular axes.

11 / UNDERSTAND & APPLY

11. Original practice with branch and degree checks

Identify the positive squared term first. For a tangent claim, verify both the contact and whether the substituted equation has retained its expected degree.

12 / UNDERSTAND & APPLY

12. The signs determine the geometry

The hyperbola has a distance difference, c²=a²+b² and eccentricity above one. Its positive quadratic term determines the transverse direction. Neither denominator is required to be the larger one, and conjugate-axis guide points are not points on the curve.

Asymptotes pass through the center and describe tail behavior, but are not tangent lines at finite points. In a line-intersection calculation, inspect the leading coefficient before interpreting a discriminant. A one-point intersection can be transverse when the line is parallel to an asymptote.

For focal distances, keep track of the branch and the order of subtraction. For parametrizations, record omitted parameter values and any omitted points. For chord formulas, verify that two real endpoints actually exist.

Finally check each formula against the defining equation. Contact points must lie on the curve, normals must be perpendicular to tangents, and asymptote slopes must match the homogeneous quadratic part. These checks distinguish genuine geometry from formally similar ellipse formulas.