Did you know that one straight line can be tangent to two different parabolas at two different points? Instead of finding their intersection, require the proposed line to share the gradient and height of each parabola at its own contact point. For two equally curved, horizontally and vertically shifted parabolas, this often gives a surprisingly simple exact answer.
For example, let f(x)=x² and g(x)=(x−4)²+8. Seek a common tangent y=mx+b. The tangent to f with gradient m has intercept −m²/4. The tangent to g with the same gradient has intercept 8−4m−m²/4. Equality gives m=2 and b=−1. Thus the common tangent is y=2x−1. It touches f at (1,1) and g at (5,9), even though the two parabolas themselves intersect at (3,9). This guide derives the result using derivatives and discriminants, verifies the distinct contacts and explains which special cases have no unique common tangent.
What a common tangent actually means
A tangent line to y=f(x) at x=t passes through (t,f(t)) and has gradient f′(t), when the derivative exists.
A common tangent to two curves must satisfy this local tangency condition for each curve, possibly at different values of x. The line is one geometric object, but its two contacts need not coincide.
This is different from finding where the curves intersect. A shared intersection is simply a point belonging to both graphs; their local tangent gradients at that point may be unequal.
Before starting, introduce two contact inputs or an unknown line gradient. Do not assume that a common tangent must pass through an intersection of the original two parabolas.
Read the two parabolas as transformations
The first curve is f(x)=x² with vertex (0,0). The second is g(x)=(x−4)²+8 with vertex (4,8).
They have the same upward curvature: the coefficient of the squared term is 1 in both. The second is obtained by moving the first four units right and eight units up.
Because their shapes agree, the tangent at a point horizontally shifted by four units has the same gradient. The accompanying tangent line is also shifted by the geometric translation.
A line common to both must be positioned so that these shifted tangent lines coincide. This observation motivates comparing their intercepts after fixing a shared gradient m.
Introduce the unknown common tangent
Write the desired line as y=mx+b, where m is its unknown gradient and b its y-intercept.
We will separately determine what b must be for this line to be tangent to f and to g.
Both parabolas are graphs of differentiable functions with finite gradients at every real x, so their tangents are not vertical. The form y=mx+b is therefore sufficient here.
At the end, one pair (m,b) must satisfy both tangency conditions. Finding two unrelated tangent equations, one for each parabola, is only an intermediate step unless they are shown to be the same line.
Find the f-contact input from its gradient
Differentiate f(x)=x². The derivative is f′(x)=2x.
If a tangent to f has gradient m, its contact input t satisfies 2t=m, so t=m/2.
The point of contact is therefore (m/2,m²/4). Apply point–slope form to get y−m²/4=m(x−m/2).
Simplify to y=mx−m²/4. Thus every tangent to the first parabola with gradient m has intercept b_f=−m²/4. This relationship is exact and avoids needing to solve a new quadratic for the contact coordinates each time.
Find the g-contact input with the same gradient
Differentiate g(x)=(x−4)²+8. Its derivative is g′(x)=2(x−4).
If its tangent has the same gradient m, the contact input u must satisfy 2(u−4)=m, so u=4+m/2.
Its height is g(u)=(m/2)²+8=m²/4+8.
Thus the second contact is (4+m/2,m²/4+8). The contact inputs differ by four, matching the horizontal shift between the parabolas. Their heights differ by eight, matching the vertical shift.
Write the g-tangent in line form
Use the point (4+m/2,8+m²/4) and gradient m to write y−(8+m²/4)=m[x−(4+m/2)].
Expand: y=mx−4m−m²/2+8+m²/4.
Combine the squared terms to obtain y=mx+8−4m−m²/4.
The intercept for this tangent is b_g=8−4m−m²/4. Its slope is already m by construction.
A common tangent requires the same intercept as the first parabola’s tangent because two straight lines with equal gradient coincide exactly when their intercepts are equal.
Equate the two tangent intercepts
The first intercept is b_f=−m²/4. The second is b_g=8−4m−m²/4.
Set b_f=b_g. The −m²/4 terms cancel, leaving 8−4m=0.
Therefore 4m=8 and m=2. Substitute into b_f to get b=−2²/4=−1.
The unique common tangent in this translated-parabola setting is y=2x−1. The cancellation is not a coincidence: the two parabolas have the same quadratic coefficient, so their tangent intercept formulas contain the same squared-gradient term.
Recover the two contact points
For f, the contact input is t=m/2=1, and f(1)=1. The first contact is T₁(1,1).
For g, the contact input is u=4+m/2=5, and g(5)=(5−4)²+8=9. The second contact is T₂(5,9).
Both points lie on y=2x−1: at x=1 the line gives 1, and at x=5 it gives 9.
The line therefore passes through both curve contacts. But line membership alone does not prove tangency; the gradients at those contacts must also be checked.
Verify both tangent gradients
At T₁, the derivative of f is f′(1)=2. This equals the line’s gradient 2.
At T₂, the derivative of g is g′(5)=2(5−4)=2. This also equals the line’s gradient.
Therefore the line y=2x−1 is tangent to both curves at their respective points.
This is a complete calculus verification: each contact lies on its parabola, lies on the line and has matching line–curve gradient. The exact checks establish the geometric claim independently of the earlier intercept algebra.
The original parabolas meet somewhere else
Solve f(x)=g(x): x²=(x−4)²+8.
Expand the right side to x²−8x+24. Cancelling x² gives 8x=24, so x=3 and y=f(3)=9. The curves intersect at (3,9).
But their gradients at x=3 are f′(3)=6 and g′(3)=2(3−4)=−2. They cross with different tangent directions.
The common tangent y=2x−1 gives y=5 at x=3, not 9. Thus it does not pass through the parabola intersection. This comparison exposes the common misconception that a common tangent must touch both graphs at a point where the graphs meet.
Prove tangency using a repeated intersection
A straight line is tangent to a parabola when substituting the line into the curve equation produces a repeated real root for the contact input.
For f(x)=x² and line y=2x−1, equate x²=2x−1. Rearranging gives x²−2x+1=(x−1)²=0.
The repeated root x=1 identifies the first contact (1,1).
For g, equate (x−4)²+8=2x−1. Rearrangement gives x²−10x+25=(x−5)²=0. The repeated root x=5 identifies (5,9). Both discriminants are zero, confirming tangency by a method independent of differentiation.
The discriminant approach starts from one line
Start with the unknown line y=mx+b and demand tangency to f(x)=x².
Substitution gives x²−mx−b=0. Its discriminant is m²+4b. Require zero for one repeated intersection, giving b=−m²/4.
Now impose tangency to g(x)=(x−4)²+8. Substituting y=mx+b produces x²−(8+m)x+(24−b)=0.
Its discriminant is (8+m)²−4(24−b)=m²+16m−32+4b. Setting it to zero and substituting b=−m²/4 gives 16m−32=0, hence m=2. This reproduces the common line.
Why the two methods agree
The calculus method requires equal tangent gradient and equal contact height along one line. The discriminant method requires the line to have one repeated intersection with each parabola.
For smooth quadratic graphs, these are two descriptions of the same tangency condition.
One method may be more efficient depending on the question. If derivatives have already been introduced, gradient matching is direct. If the paper is focused on quadratic roots and discriminants, repeated intersections can be convenient.
A student should choose one complete method and use the other as a check when time permits. Mixing pieces of both methods without finishing either can create unnecessary algebra.
A general horizontal and vertical shift
Let f(x)=x² and g(x)=(x−h)²+k, where h and k are real constants and h≠0.
A tangent to f with gradient m is y=mx−m²/4. The corresponding tangent to g has the same gradient and equation y=mx−hm−m²/4+k.
For one common line, their intercepts must match. Thus −hm+k=0, giving m=k/h.
The common tangent is therefore y=(k/h)x−k²/(4h²). Its contact inputs are t=k/(2h) for f and u=h+k/(2h) for g. This formula depends on the equal curvature and nonzero horizontal shift assumptions.
Verify the general contact pair
For the first parabola, t=m/2 and f(t)=m²/4. The proposed line y=mx−m²/4 evaluated at x=m/2 also gives m²/4.
For the shifted second parabola, u=h+m/2 and g(u)=k+m²/4.
The common line at u has value m(h+m/2)−m²/4=mh+m²/4. Since mh=k, this equals k+m²/4.
The derivatives at the two contacts are 2t=m and 2(u−h)=m. Thus the general formula passes both the point and gradient checks and is not merely an algebraic guess based on the numerical example.
A horizontal shift with no vertical shift
Suppose k=0 but h≠0. Then g(x)=(x−h)² is the original parabola shifted horizontally with no change in vertex height.
The common-tangent slope is m=k/h=0. The common line becomes y=0.
It touches f at (0,0) and g at (h,0), the two vertices. Both tangent gradients are zero.
For example, y=x² and y=(x−4)² share the horizontal tangent y=0 even though their vertices occur at different x-coordinates. This special case is a useful check on the general formula when the vertical offset vanishes.
What if the parabolas have only a vertical shift?
Let h=0, so g(x)=x²+k. The graphs have the same axis of symmetry and are vertically separated by k.
A tangent to f of gradient m has intercept −m²/4. A tangent to g with that same gradient has intercept k−m²/4.
If k≠0, these intercepts can never be equal. Therefore no straight line is tangent to both vertically separated copies at possibly different points.
If k=0, the two curves coincide and every tangent of the shared parabola is common. This explains why the formula m=k/h cannot be used when h=0: division by zero hides two qualitatively different cases.
Generalise to another nonzero curvature coefficient
Let f(x)=ax² and g(x)=a(x−h)²+k, where a≠0 and h≠0. These parabolas have the same coefficient a, though their opening direction depends on its sign.
A tangent to f with gradient m has contact input t=m/(2a) and line intercept −m²/(4a).
The same-gradient tangent to g has intercept −hm−m²/(4a)+k.
Equal intercepts again require hm=k, so m=k/h. The common tangent is y=(k/h)x−k²/(4ah²). The equal curvature allows the squared-gradient terms to cancel just as in the monic case.
A second worked example with a steeper parabola
Consider f(x)=2x² and g(x)=2(x−3)²+6. Here a=2, h=3 and k=6.
The common tangent gradient is m=k/h=2. Its intercept is −m²/(4a)=−4/8=−1/2.
Therefore the line is y=2x−1/2. It touches f at t=m/(2a)=1/2, where f(t)=1/2.
The second contact is u=h+t=7/2. Its height is g(7/2)=2(1/2)²+6=13/2. The line also gives 2(7/2)−1/2=13/2, and both derivatives equal 2. Every condition checks.
The common tangent can slope downward
In the formula m=k/h, the slope is negative when the horizontal and vertical shifts have opposite signs.
For example, take f(x)=x² and g(x)=(x−4)²−8. Now h=4 and k=−8, so m=−2.
The common tangent intercept is −m²/4=−1, giving y=−2x−1.
It touches f at t=−1, where f(−1)=1 and the line also gives 1. It touches g at u=3, where g(3)=1−8=−7 and the line gives −7. The slope sign follows the translation direction, not the fact that both parabolas open upward.
The shared tangent may have widely separated contacts
For equal-curvature translated parabolas, the contact inputs differ by h, the horizontal shift.
In the main example, the inputs are 1 and 5, four units apart. For a larger |h|, the points of tangency can be farther apart, even when the gradient m remains modest.
The corresponding contact heights differ by k because the two parabolas are vertical translations as well.
This is a geometric invariant of the construction: the vector between corresponding contact points is (h,k). The common tangent must be parallel to that translation vector, which explains why its gradient is k/h when h≠0.
A geometric interpretation of m=k/h
The displacement from the first contact point to the second is (h,k), as the second parabola is a translation of the first.
Both contact points lie on one common straight line. Therefore the gradient of that line equals vertical displacement divided by horizontal displacement: m=k/h.
This provides a short geometric explanation of the algebraic result. The shared tangent is parallel to the vector that translates the first parabola onto the second.
However, this reasoning relies on the contacts corresponding under the translation, which follows from equal gradients and equal curvature. A pair of parabolas with different quadratic coefficients needs a fresh analysis rather than an unproved copy of the same formula.
A parameter can be recovered from the common tangent slope
Suppose f(x)=x² and g(x)=(x−4)²+k have a common tangent with gradient 3.
The general relationship is m=k/h. With h=4 and m=3, the required vertical shift is k=12.
The common tangent to f with slope 3 has equation y=3x−9/4. It touches f at x=3/2.
For the shifted parabola, the corresponding contact input is 4+3/2=11/2. Its output is (3/2)²+12=57/4, while the line gives 3(11/2)−9/4=66/4−9/4=57/4. This confirms the recovered parameter.
Why common tangency does not require curve equality
Setting f(x)=g(x) finds shared graph coordinates. The main parabolas have one such intersection at (3,9), but there their derivatives are 6 and −2.
Setting their derivatives equal at the same x would give 2x=2(x−4), or 0=−8, so these two shifted copies cannot have matching slopes at the same input.
Nevertheless, the common tangent touches f at x=1 and g at x=5. Equal gradients occur at different x-inputs.
A common tangent problem should therefore introduce two contact inputs or one shared gradient, not demand f(t)=g(t) at the same t unless the wording explicitly requires tangency at a common intersection.
What changes if the quadratic coefficients are unequal?
For f(x)=ax² and g(x)=b(x−h)²+k with a and b nonzero but a≠b, equal tangent gradient m corresponds to contact inputs m/(2a) and h+m/(2b).
The tangent intercepts are −m²/(4a) and k−hm−m²/(4b). Equating them gives a quadratic equation in m in general.
That quadratic may have two, one or no real solutions depending on the coefficients. The uniqueness of the equal-curvature shifted case no longer follows.
This extension connects common tangents with discriminants and parameter root counts. It is appropriate for advanced practice after the student has mastered the two equal-curvature examples and the meaning of a repeated line–parabola intersection.
A concise method for a school solution
Start with an unknown common gradient m. Find the first curve’s contact input by solving f′(t)=m and write its tangent in y=mx+b form.
Repeat for the second curve using g′(u)=m. Equate the two intercepts to obtain a common line.
Solve for m, find b, and state the common tangent equation. Then calculate both contact points and check that the line passes through them with gradients matching the two derivatives.
If the question uses discriminants rather than calculus, require a repeated root in each line–curve intersection equation. Keep the contact-point distinction visible throughout the solution.
Practice questions with checked answers
Question one: f(x)=x² and g(x)=(x−4)²+8. Common tangent: y=2x−1; contacts (1,1) and (5,9).
Question two: f(x)=x² and g(x)=(x−4)². Common tangent: y=0; contacts at their vertices.
Question three: f(x)=x² and g(x)=(x−4)²−8. Common tangent: y=−2x−1.
Question four: f(x)=2x² and g(x)=2(x−3)²+6. Common tangent: y=2x−1/2.
Question five: for f=x² and g=(x−h)²+k with h≠0, find the common gradient. Answer: m=k/h.
Question six: if g=x²+5 while f=x², how many common tangents exist? None, because the curves are distinct vertical translations.
For every answer, confirm the line is tangent to both parabolas at its own contact, not merely that the line intersects both.
The common line lies below both upward parabolas
For the opening example, compare the first curve with the common tangent: f(x)−(2x−1)=x²−2x+1=(x−1)².
This is non-negative for every real x, with equality only at x=1. Thus the line y=2x−1 lies below or on the first parabola everywhere and touches it exactly at (1,1).
For the second curve, g(x)−(2x−1)=(x−4)²+8−2x+1=(x−5)².
This is also non-negative for every real x, with equality only at x=5. Therefore the common tangent is a supporting line beneath both upward-opening parabolas, not a line that crosses either curve at its contact point.
The squared-gap identities give an immediate verification
The identities f(x)−ℓ(x)=(x−1)² and g(x)−ℓ(x)=(x−5)² verify more than two points on a line.
They prove that the line is never above either parabola and has exactly one contact with each, because a real square vanishes only when its bracket is zero.
The corresponding repeated intersection roots are visible without another discriminant computation. The line meets f at x=1 and g at x=5, each with multiplicity two in the difference equation.
This is a useful secondary proof of tangency and connects the problem to quadratic inequalities. It also shows the special importance of complete-square forms in recognising a tangent as a global supporting line.
The contact-to-contact segment lies along the common tangent
The two contact points are T₁(1,1) and T₂(5,9). Their difference vector is (4,8), with gradient 8/4=2.
That agrees with the common line gradient m=2. The distance between the contacts is √(4²+8²)=√80=4√5 units.
The segment joining them belongs entirely to the common tangent line, although its endpoints lie on different parabolas.
This is a geometric check independent of differentiation: if two purported contact points do not lie on the same candidate line, their joining vector cannot have the common tangent’s direction. The distance computation can also be useful when a question extends the tangent problem into coordinate geometry.
General supporting-line gaps for equal-curvature shifts
For f(x)=ax² and g(x)=a(x−h)²+k with a>0 and h≠0, let m=k/h and ℓ(x)=mx−m²/(4a).
The first gap is f(x)−ℓ(x)=a[x−m/(2a)]².
Using mh=k, the second gap simplifies to g(x)−ℓ(x)=a[x−h−m/(2a)]².
Both are non-negative for every real x because a>0. They vanish exactly at the two tangent contacts separated horizontally by h.
This derivation proves that the common tangent supports both upward-opening curves globally. If a<0, the signs reverse and the line lies above both downward-opening parabolas instead. The gradient and contact formulas remain otherwise valid.
The line is fixed by the translation vector
A translation by (h,k) maps the first parabola to the second. Its tangent at a given point maps to a parallel tangent through the translated contact point.
To have one tangent line invariant under this translation, the translation vector itself must run along that line. For h≠0, that means its slope is k/h.
This gives the geometry behind the algebraic equality hm=k. The two contacts lie on one line because their displacement equals the translation vector.
The condition is particularly easy to visualise when k=0: the translation is horizontal, so the shared tangent is horizontal at the two vertices. When h=0 and k≠0, the translation is vertical and cannot preserve a finite-slope tangent line of a parabola.
A quick check for an incorrect candidate tangent
Suppose a student proposes the line y=x+1 as a common tangent to f(x)=x² and g(x)=(x−4)²+8.
For f, the tangent gradient would be 1 at x=1/2, where f(1/2)=1/4. But the proposed line gives 3/2 at that x, so it does not pass through the needed contact.
The point and gradient conditions fail immediately. The correct common gradient is 2, not 1.
This example shows why matching a line’s slope to a derivative is only half of tangency. The line must also have the correct intercept to meet the curve at that contact. For two curves, both sets of conditions must hold simultaneously.
Syllabus scope and connected reading
Common tangents between curves combine differentiation, simultaneous algebra, quadratic discriminants, graphs and geometric transformations. Two-parabola common tangency can be useful enrichment beyond routine tangent-to-one-curve questions, so match the depth to the learner’s syllabus and teacher guidance.
For Singapore, see SEAB’s relevant 2026 O-Level Additional Mathematics and 2027 SEC G3 syllabus lists.
Related guides include How to Find the Angle Between Two Curves at an Intersection, How to Recover Unknown Coefficients from a Tangent Line and How to Prove a Quadratic Stays Above a Line for Every x. More resources are available in the Additional Mathematics archive and the eduKate small-group reference.
Parents may enquire about suitable eduKate Singapore three-student tutorials near Sixth Avenue MRT. The enduring skill is to understand tangency as a local agreement in gradient and height, then apply that condition consistently to two distinct contact points. Properly taught kids shine a bright light into the future.