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Conic Sections: Star problems

Ten optional challenges to stretch your reasoning. Work on paper, use hints when you need them, and check the answer or full solution when you are ready. You can skip these problems and continue the course.

1 of 3 stars: Stretch
2 of 3 stars: Challenge
3 of 3 stars: Deep challenge

Stars indicate difficulty within this set.

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Problem 1 of 10

Problem 1 A circle through two crossings

Difficulty: 1 of 3 stars, Stretch

The circles $x^2+y^2=25$ and $(x-6)^2+y^2=13$ meet at two points $A$ and $B$ . Find every circle through both $A$ and $B$ that is tangent to the $y$ -axis. Give its center and radius, and prove that your list is complete.

Text description of this figure

Coordinate axes. A larger circle is centered at the origin, which is labeled O. A smaller circle is centered at the marked point (6, 0) on the positive x-axis. The two circles cross at two points: A above the x-axis and B below it, mirror images of each other in the x-axis.

Builds on Circles
1. Hint 1
  Subtract the two circle equations before solving for either intersection.
2. Hint 2
  A center equidistant from $A$ and $B$ lies on their perpendicular bisector. Tangency to the $y$ -axis makes the radius equal to the absolute value of the center's horizontal coordinate.
Answer

Exactly one circle: center $(25/8,0)$ and radius $25/8$ .

Full solution

Subtracting the first equation from the second gives $-12x+36=-12$ , hence $x=4$ .
Substitution gives $y^2=9$ , so $A=(4,3)$ and $B=(4,-3)$ .

Write the center of any circle through these points as $(h,k)$ .
Equality of its squared distances to $A$ and $B$ gives $(k-3)^2=(k+3)^2$ , forcing $k=0$ .
Its squared radius is therefore
$r^2=(h-4)^2+9.$
This describes every possible circle through the crossings, because either distance determines a positive radius.

The distance from $(h,0)$ to the $y$ -axis is $|h|$ .
Tangency is therefore equivalent to $r^2=h^2$ , which gives $h^2=(h-4)^2+9$ , or $8h=25$ .
Thus $h=25/8$ and $r=25/8$ .
These values satisfy both point conditions and the tangency condition.
The center is not assumed to lie on a particular side of the axis; the equation has already exhausted both possibilities.

Answer

Exactly one circle: center $(25/8,0)$ and radius $25/8$ .

Key idea

Subtracting equal-degree distance equations often turns a family of circles into a one-variable problem.
Problem 1: Mark complete
Problem 2 A broken path through a parabola

Difficulty: 1 of 3 stars, Stretch

A parabola has focus $F=(0,2)$ and directrix $y=-2$ . A point $P$ may be anywhere on the parabola, and $A=(6,5)$ is fixed. Find the least possible value of $PF+PA$ , and find every point $P$ that attains it. Justify the minimum geometrically or algebraically.

Text description of this figure

Coordinate axes with an upward-opening parabola whose lowest point is at the origin. The focus F, labeled (0, 2), is marked on the positive y-axis. The directrix, the horizontal line y equals negative 2, is drawn dashed below the x-axis. A point P on the right arm of the parabola, level with F, is joined to F by a horizontal segment and to the fixed point A, labeled (6, 5), by a second segment, making a broken path from F to P to A. The point A lies above the right arm of the parabola, inside its cup.
1. Hint 1
  Replace $PF$ by the distance from $P$ to the directrix.
2. Hint 2
  If $P=(x,y)$ , then $PA\ge |y-5|$ . Determine when this inequality can be an equality, and check the corresponding point on the parabola.
Answer

The minimum is $7$ , attained only at $P=(6,9/2)$ .

Full solution

For a point $P=(x,y)$ on the parabola, the focus-directrix equation is $x^2+(y-2)^2=(y+2)^2$ , so $x^2=8y$ and $y\ge0$ .
Consequently $PF=y+2$ , with no absolute-value ambiguity.

The vertical separation is at most the full distance to $A$ :
$PA\ge|y-5|.$
Hence

$PF+PA\ge y+2+|y-5|\ge7.$

The second inequality is an equality exactly when $y\le5$ .
The first is an equality exactly when $x=6$ , since
$PA^2=(x-6)^2+(y-5)^2.$
On the parabola, $x=6$ forces $y=36/8=9/2$ , which does satisfy $y\le5$ .
At this point $PF=13/2$ and $PA=1/2$ , so the lower bound is attained.

Every equality condition has been retained, so there is no second minimizing point.
The useful change was to replace one slanted segment by a vertical distance using the geometric definition of the parabola.

Answer

The minimum is $7$ , attained only at $P=(6,9/2)$ .

Key idea

When a conic is defined by distances, use that definition before replacing everything by coordinates.
Problem 2: Mark complete
Problem 3 A triangle with a tilted fixed base

Difficulty: 1 of 3 stars, Stretch

An ellipse consists of points $P=(x,y)$ whose distances to $F_1=(-3,0)$ and $F_2=(3,0)$ add to $10$ . Two other points, $A=(0,2)$ and $B=(5,-1)$ , are fixed. Find the greatest possible area of triangle $APB$ , and determine every maximizing point $P$ .

You may use the coordinate-area fact that this triangle has area $|3x+5y-10|/2$ . Prove a bound valid for the entire ellipse, and check its equality cases.

Text description of this figure

Coordinate axes with an ellipse centered at the origin, wider than it is tall. Its foci, F1 on the negative x-axis and F2 on the positive x-axis, are marked at equal distances from the origin. A triangle joins three points: A, labeled (0, 2), on the positive y-axis inside the ellipse; B, labeled (5, negative 1), near the right end of the ellipse just below the x-axis; and P, a point on the upper left part of the ellipse.
1. Hint 1
  First obtain the ellipse equation from its focal distances. The base is tilted, so maximizing $|y|$ alone is insufficient.
2. Hint 2
  Put $U=x/5$ and $V=y/4$ . Then $U^2+V^2=1$ . Expand $(15U+20V)^2+(20U-15V)^2$ .
Answer

The maximum area is $35/2$ , attained only at $P=(-3,-16/5)$ .

Full solution

Let $r_1=PF_1$ and $r_2=PF_2$ .
We have $r_1+r_2=10$ and $r_1^2-r_2^2=12x$ , so $r_1-r_2=6x/5$ and
$r_1=5+3x/5.$
Squaring and using $r_1^2=(x+3)^2+y^2$ gives
$x^2/25+y^2/16=1.$

Put $U=x/5$ and $V=y/4$ , so $U^2+V^2=1$ .
The square identity

$(15U+20V)^2+(20U-15V)^2=625(U^2+V^2)=625$

shows that
$-25\le3x+5y\le25.$
Consequently $-35\le3x+5y-10\le15$ , and the triangle area is at most $35/2$ .
Notice that the negative endpoint, not the positive endpoint, produces the larger absolute value.

Equality requires $15U+20V=-25$ and $20U-15V=0$ .
Solving gives $U=-3/5$ , $V=-4/5$ , hence
$P=(-3,-16/5).$
At this point the two focal distances are $16/5$ and $34/5$ , which sum to $10$ , so the point belongs to the original ellipse.
Its area is
$|-9-16-10|/2=35/2.$
All equality conditions force this single point, proving uniqueness without introducing extraneous points through squaring.

Answer

The maximum area is $35/2$ , attained only at $P=(-3,-16/5)$ .

Key idea

For a tilted geometric target, bound the corresponding linear expression on the conic and retain both signs until the end.
Problem 3: Mark complete
Problem 4 Integer points beside an asymptote

Difficulty: 2 of 3 stars, Challenge

Consider the part of the hyperbola $x^2-y^2=96$ with $x>0$ and $y\ge0$ .

(a) Find all points on this part whose coordinates are integers, and find the least possible value of $x-y$ among them.

(b) If real coordinates are allowed instead, does $x-y$ have a least value? Prove your answer and explain its connection to the line $y=x$ .
1. Hint 1
  For integer coordinates, write $(x-y)(x+y)=96$ and keep track of parity.
2. Hint 2
  For real coordinates, prescribe a positive value $d=x-y$ . Solve for $x+y$ and check when the reconstructed point has $y\ge0$ .
Answer

(a) $(25,23),(14,10),(11,5),(10,2)$ ; the least value is $2$ . (b) No least value exists; $x-y$ can approach $0$ from above.

Full solution

Put $a=x-y$ and $b=x+y$ .
Because $x^2-y^2=96>0$ , we have $0<a\le b$ , $ab=96$ , and, for integer coordinates, $a$ and $b$ have the same parity.
Both must be even.
Also
$a\le\sqrt{96}<10.$
The even divisors of $96$ below $10$ are $2,4,6,8$ , and their partners $48,24,16,12$ are all even.

Recovering $x=(a+b)/2$ and $y=(b-a)/2$ gives exactly $(25,23),(14,10),(11,5),(10,2)$ .
The finite bound on $a$ proves completeness.
The minimum of $x-y=a$ is $2$ .

For real coordinates, choose any $d$ with $0<d\le\sqrt{96}$ and set $x=(d+96/d)/2$ ,
$y=(96/d-d)/2.$
These values satisfy the equation, have the required signs, and give $x-y=d$ .
Therefore arbitrarily small positive values occur, but $0$ cannot occur because it would make $x^2-y^2=0$ .

The perpendicular distance to $y=x$ is $(x-y)/\sqrt2$ .
Thus real points approach this asymptote without reaching it; the integer points have a positive minimum separation.

Answer

(a) $(25,23),(14,10),(11,5),(10,2)$ ; the least value is $2$ . (b) No least value exists; $x-y$ can approach $0$ from above.

Key idea

An integer restriction can turn a continuous family with no minimum into a finite problem with an attained minimum.
Problem 4: Mark complete
Problem 5 Midpoints of parallel chords

Difficulty: 2 of 3 stars, Challenge

A chord of the ellipse $x^2/25+y^2/9=1$ joins two distinct points, and its supporting line has slope $3/5$ .

(a) Find the exact locus of its midpoint, including which endpoints of the locus are excluded.

(b) Find all such chords whose length is $\sqrt{34}$ , giving their endpoints.

Text description of this figure

Coordinate axes with an ellipse centered at the origin, wider than it is tall. A chord, labeled slope 3 over 5, rises from the lower left part of the ellipse to the upper right part, and its two endpoints on the ellipse are marked. Its midpoint, labeled M equals h, k, is marked on the chord a little to the right of the y-axis and a little below the x-axis.
1. Hint 1
  If the midpoint is $(h,k)$ , express the endpoints as $(h+5t,k+3t)$ and $(h-5t,k-3t)$ , with $t>0$ .
2. Hint 2
  Subtract the two ellipse equations to find a midpoint relation, then add them to relate $t$ to the midpoint. The chord length is $2t\sqrt{34}$ .
Answer

(a) The open segment $3h+5k=0$ , $|h|<5/\sqrt2$ . (b) Endpoints $(5,0),(0,-3)$ , or endpoints $(-5,0),(0,3)$ .

Full solution

Write the endpoints as $(h\pm5t,k\pm3t)$ with $t>0$ .
Subtracting their ellipse equations yields
$4t(h/5+k/3)=0.$
Thus $3h+5k=0$ , or $k=-3h/5$ .
Averaging the equations then gives

$\frac{h^2}{25}+\frac{k^2}{9}+2t^2=1,\qquad t^2=\frac12-\frac{h^2}{25}.$

Because the endpoints must be distinct, $t>0$ , so
$|h|<5/\sqrt2.$
Conversely, any $h$ in this open interval determines $k=-3h/5$ and a positive $t$ by the displayed formula.
Substitution shows that both endpoints lie on the ellipse.
This proves the full locus, not only a necessary equation.
At the two excluded endpoints the chord collapses to a single point.

The endpoint displacement is $(10t,6t)$ , so the chord length is $2t\sqrt{34}$ .
Length $\sqrt{34}$ forces $t=1/2$ , and then $h^2=25/4$ .
For $(h,k)=(5/2,-3/2)$ the endpoints are $(5,0),(0,-3)$ ; for $(-5/2,3/2)$ they are $(-5,0),(0,3)$ .
Both have the stated slope and length.

Answer

(a) The open segment $3h+5k=0$ , $|h|<5/\sqrt2$ . (b) Endpoints $(5,0),(0,-3)$ , or endpoints $(-5,0),(0,3)$ .

Key idea

Expressing two endpoints as midpoint plus or minus a displacement separates position from direction.
Problem 5: Mark complete
Problem 6 A chord through the focus

Difficulty: 2 of 3 stars, Challenge

The parabola $y^2=12x$ has focus $F=(3,0)$ . A line through $F$ meets the parabola at two distinct finite points $P$ and $Q$ .

(a) Prove that $1/PF+1/QF=1/3$ .

(b) Find the least possible chord length $PQ$ , and determine every chord attaining it. Your argument must include vertical lines and explain why the horizontal line is excluded.
1. Hint 1
  The directrix is $x=-3$ , so a point $(x,y)$ on the parabola has focal distance $x+3$ .
2. Hint 2
  For a nonvertical line write $y=m(x-3)$ . Use the sum and product of its two intersection inputs. Separately check $x=3$ .
Answer

(a) The reciprocal sum is always $1/3$ . (b) The least length is $12$ , attained only by the chord from $(3,-6)$ to $(3,6)$ .

Full solution

The horizontal line through $F$ is $y=0$ , which meets the parabola only at its vertex, so it is not allowed.
For a nonvertical allowed line, $y=m(x-3)$ with $m\ne0$ .
Substitution gives
$m^2x^2-(6m^2+12)x+9m^2=0.$
If the intersection inputs are $u,v$ , their product is $9$ and their sum is $6+12/m^2$ .

The focal distances are $u+3$ and $v+3$ .
Therefore

$\frac1{u+3}+\frac1{v+3}=\frac{u+v+6}{uv+3(u+v)+9}=\frac13.$

To relate these distances to chord length, put $x=3+s$ .
The intersection equation is $m^2s^2-12s-36=0$ , whose two roots have opposite signs.
Hence $F$ lies between $P$ and $Q$ , and
$PQ=PF+QF=u+v+6=12+12/m^2>12.$

The remaining line is vertical: $x=3$ gives $P=(3,-6)$ and $Q=(3,6)$ .
Each focal distance is $6$ , so its reciprocal sum is also $1/3$ and its chord length is $12$ .
Thus the vertical chord is the unique minimizer.
It cannot be omitted merely because a slope parametrization misses it.

Answer

(a) The reciprocal sum is always $1/3$ . (b) The least length is $12$ , attained only by the chord from $(3,-6)$ to $(3,6)$ .

Key idea

A root sum and product can reveal a geometric invariant, but omitted line directions need their own check.
Problem 6: Mark complete
Problem 7 Circles between two boundaries

Difficulty: 2 of 3 stars, Challenge

A fixed circle has center $(0,3)$ and radius $3$ . A second circle has positive radius, lies in the closed upper half-plane, and is tangent to the $x$ -axis. It is externally tangent to the fixed circle, meaning the distance between their centers is the sum of their radii.

(a) Find the locus of the second circle's center.

(b) Find all such circles that are also tangent to the line $x=6$ . Give their centers and radii, and check that the contacts are external as required.

Builds on Circles
1. Hint 1
  If the second center is $(u,v)$ , its radius must be $v>0$ . Translate external tangency into an equation of distances.
2. Hint 2
  After squaring, the center locus is a parabola. Tangency to $x=6$ adds $v=|u-6|$ ; solve both sign cases.
Answer

(a) $v=u^2/12$ , with $u\ne0$ . (b) Centers $(-6\pm6\sqrt3,\,12\mp6\sqrt3)$ , with corresponding radii $12\mp6\sqrt3$ .

Full solution

Tangency to the $x$ -axis and location above it make the radius $v$ , where $v>0$ .
External tangency to the fixed circle is

$\sqrt{u^2+(v-3)^2}=v+3.$

Both sides are positive, so squaring is reversible.
Simplifying gives $u^2=12v$ .
The condition $v>0$ excludes $u=0$ .
Conversely every remaining point of this parabola gives a positive-radius circle satisfying the original distance condition; its lowest point has height zero.

Tangency to $x=6$ requires $v=|u-6|$ .
If $u\ge6$ , then $u^2=12u-72$ , or $(u-6)^2+36=0$ , which has no real solution.
If $u<6$ , then $u^2=72-12u$ , giving
$(u+6)^2=108.$
Hence $u=-6\pm6\sqrt3$ and
$v=6-u=12\mp6\sqrt3.$
Both horizontal coordinates are less than $6$ , and both radii are positive.

For each candidate, $u^2=12v$ implies that the center distance is exactly $v+3$ , rather than $|v-3|$ .
Thus the two circles touch externally, and all original tangency and half-plane requirements hold.

Answer

(a) $v=u^2/12$ , with $u\ne0$ . (b) Centers $(-6\pm6\sqrt3,\,12\mp6\sqrt3)$ , with corresponding radii $12\mp6\sqrt3$ .

Key idea

A family of tangent circles can be encoded by a conic traced out by their centers.
Problem 7: Mark complete
Problem 8 One meeting point is not always tangency

Difficulty: 3 of 3 stars, Deep challenge

Find every line through $(0,3)$ that meets the hyperbola $x^2/16-y^2/9=1$ in exactly one distinct real point. Give the meeting point for each line.

For each answer, determine whether substitution into the hyperbola produces a quadratic with a repeated root or a linear equation. Explain geometrically why the latter possibility occurs. Include the vertical line in your completeness check.

Text description of this figure

Coordinate axes with a hyperbola whose two branches open to the left and to the right. The branches cross the x-axis at their vertices, labeled negative 4 and 4. The point (0, 3) is marked on the positive y-axis, between the two branches. No asymptotes are drawn.
1. Hint 1
  Substitute $y=mx+3$ , but do not apply the discriminant until you know the coefficient of $x^2$ is nonzero.
2. Hint 2
  The coefficient of $x^2$ vanishes when $m$ equals an asymptote slope. These exceptional slopes can still produce a single intersection.
Answer

Repeated-root lines: $y=3\pm(3\sqrt2/4)x$ , with meeting points $(\mp4\sqrt2,-3)$ . Linear cases: $y=3+(3/4)x$ at $(-4,0)$ and $y=3-(3/4)x$ at $(4,0)$ .

Full solution

The vertical line through $(0,3)$ is $x=0$ , which gives $-y^2/9=1$ and has no real meeting point.
Every other line is $y=mx+3$ .
Substitution and multiplication by $144$ give

$(9-16m^2)x^2-96mx-288=0.$

If $m\ne\pm3/4$ , this is a genuine quadratic.
Its discriminant is $10368-9216m^2$ , which vanishes exactly at
$m=\pm3\sqrt2/4.$
The repeated input is $x=-16m/3$ , yielding points $(\mp4\sqrt2,-3)$ .
These are the two tangent lines detected by repeated roots.

If $m=3/4$ , the equation instead becomes $-72x-288=0$ , giving $(-4,0)$ .
If $m=-3/4$ , it becomes $72x-288=0$ , giving $(4,0)$ .
The line slopes $\pm3/4$ are the hyperbola's asymptote slopes.
Along these directions, cancellation removes the quadratic term; a translated asymptote can meet the hyperbola once without being a tangent.

All four listed points satisfy their original line and conic equations.
The vertical line, the two degenerate coefficients, and every genuine quadratic case have now been accounted for.

Answer

Repeated-root lines: $y=3\pm(3\sqrt2/4)x$ , with meeting points $(\mp4\sqrt2,-3)$ . Linear cases: $y=3+(3/4)x$ at $(-4,0)$ and $y=3-(3/4)x$ at $(4,0)$ .

Key idea

The degree of an intersection equation can change with a parameter; a discriminant test alone may miss valid cases.
Problem 8: Mark complete
Problem 9 Two conics with the same foci

Difficulty: 3 of 3 stars, Deep challenge

The ellipse $x^2/25+y^2/16=1$ and hyperbola $x^2-y^2/8=1$ have the same two foci.

(a) Find the foci and every intersection point.

(b) At an intersection $(u,v)$ , show by algebra that the lines $ux/25+vy/16=1$ and $ux-vy/8=1$ each meet the corresponding conic only at $(u,v)$ . Then prove that these two contact lines are perpendicular. Do not use calculus.

Text description of this figure

Coordinate axes with an ellipse centered at the origin, wider than it is tall, and a hyperbola whose two branches open to the left and to the right from vertices close to the origin. Each branch rises and falls steeply through the ellipse, crossing it once above the x-axis and once below it. The curves are labeled ellipse and hyperbola, and the crossing points are not marked.
1. Hint 1
  Treat $x^2$ and $y^2$ as the unknowns in part (a).
2. Hint 2
  For part (b), set $x=u+h$ and use each line to express $y-v$ in terms of $h$ . The constant and linear terms in the conic equation cancel.
Answer

The common foci are $(\pm3,0)$ , and the four intersections are $(\pm5/3,\,\pm8\sqrt2/3)$ , with signs chosen independently. The two displayed contact lines are perpendicular at each intersection.

Full solution

The ellipse has squared focal distance $25-16=9$ , while the hyperbola has squared focal distance $1+8=9$ , so both have foci $(\pm3,0)$ .
Put $X=x^2$ and $Y=y^2$ .
The equations become $16X+25Y=400$ and $8X-Y=8$ , giving $X=25/9$ and $Y=128/9$ .
Both are positive, so all four independent sign choices yield actual intersections.

Fix any intersection $(u,v)$ ; in particular $u,v\ne0$ .
On the ellipse contact line, write $x=u+h$ .
Since $(u,v)$ lies on the line,
$y=v-16uh/(25v).$
Substitute this into the ellipse equation and subtract the equation at $(u,v)$ .
The result is

$\left(\frac1{25}+\frac{16u^2}{625v^2}\right)h^2=0.$

The coefficient is positive, so $h=0$ .
Thus this line meets the ellipse only at $(u,v)$ .
For the hyperbola line, the analogous substitution is $y=v+8uh/v$ , and subtraction gives
$(1-8u^2/v^2)h^2=0.$
Here the coefficient is $1-8(25/128)=-9/16\ne0$ , so this contact is also unique.

The slopes are $-16u/(25v)$ and $8u/v$ .
Their product is
$-128u^2/(25v^2)=-1.$
Hence the lines are perpendicular for every sign choice.
The contact equations were verified directly, so no tangent formula or derivative was assumed.

Answer

The common foci are $(\pm3,0)$ , and the four intersections are $(\pm5/3,\,\pm8\sqrt2/3)$ , with signs chosen independently. The two displayed contact lines are perpendicular at each intersection.

Key idea

Shared geometric data can produce an algebraic relation between slopes that is invisible in either conic alone.
Problem 9: Mark complete
Problem 10 The largest inscribed triangle

Difficulty: 3 of 3 stars, Deep challenge

Three distinct points on the ellipse $x^2/25+y^2/9=1$ form a triangle. Find its greatest possible area and give one maximizing triangle. Also characterize all maximizing triangles.

You may use the elementary fact that stretching every horizontal distance by $a>0$ and every vertical distance by $b>0$ multiplies every triangle's area by $ab$ . Your bound must apply to all inscribed triangles, without assuming a side is horizontal. Do not use calculus or trigonometry.

Text description of this figure

Coordinate axes with an ellipse centered at the origin, wider than it is tall. A triangle is inscribed in it, with its three vertices marked on the ellipse: one at the right end of the ellipse on the positive x-axis, one at the top on the positive y-axis, and one on the lower left part of the ellipse.
1. Hint 1
  The coordinate change $(x,y)\mapsto(x/5,y/3)$ sends the ellipse to the unit circle. Solve the triangle-area problem there first.
2. Hint 2
  For a chord of the unit circle at distance $d$ from the center, the base is $2\sqrt{1-d^2}$ and the greatest possible height is $1+d$ . Set $t=1+d$ and factor $27/16-t^3(2-t)$ .
Answer

The maximum is $45\sqrt3/4$ . One example has vertices $(5,0),(-5/2,3\sqrt3/2),(-5/2,-3\sqrt3/2)$ . Equality holds exactly when the transformed vertices $(x/5,y/3)$ form an equilateral triangle on the unit circle.

Full solution

The stated scaling takes the ellipse to the unit circle and divides all areas by $15$ .
Consider any triangle on that circle, choose any one of its sides as a base, and let $d\in[0,1)$ be the distance from the center to the base line.
Pythagoras gives base length $2\sqrt{1-d^2}$ .
The third vertex has distance at most $1+d$ from that line, with equality precisely at a circle point farthest from it; this point is unique when $d>0$ .

Thus the transformed area $K$ satisfies
$K^2\le(1-d^2)(1+d)^2.$
Put $t=1+d$ , so $1\le t<2$ .
Then

$\frac{27}{16}-t^3(2-t)=\left(t-\frac32\right)^2\left(t^2+t+\frac34\right)\ge0.$

Hence
$K\le3\sqrt3/4.$
Equality requires $d=1/2$ and the third vertex at the farthest point.
The base then has length $\sqrt3$ ; its half-length is $\sqrt3/2$ and the height is $3/2$ , so each other side also has length $\sqrt3$ .
These are exactly the equilateral triangles on the unit circle, all of which attain the bound.

Multiplying by $15$ gives maximum area $45\sqrt3/4$ and the stated equality characterization.
The displayed example is obtained by scaling the unit-circle vertices $(1,0),(-1/2,\pm\sqrt3/2)$ .
Choosing an arbitrary base made the bound independent of its direction.

Answer

The maximum is $45\sqrt3/4$ . One example has vertices $(5,0),(-5/2,3\sqrt3/2),(-5/2,-3\sqrt3/2)$ . Equality holds exactly when the transformed vertices $(x/5,y/3)$ form an equilateral triangle on the unit circle.

Key idea

A coordinate stretch can turn a difficult shape into a symmetric one while preserving a controllable area ratio.
Problem 10: Mark complete