Coordinate Geometry: Straight Lines, Circles and Conics

Parabola latus rectum

A parabola is the locus of points equidistant from a fixed point, the focus, and a fixed line, the directrix. For the standard form y^2 = 4ax with a > 0, the vertex is the origin, the focus is (a, 0), and the directrix is x = -a. The latus rectum is the focal chord perpendicular to the axis of symmetry, so for this form it is the line x = a with endpoints (a, \pm 2a). Its length is 4a, which is four times the focal distance a, not twice it.

• latus rectum length — for y^2 = 4ax or x^2 = 4ay, the length is 4|a|, and the endpoints lie at distance 2|a| on either side of the focus. e.g. For y^2 = 12x, the latus rectum length is 12.
• opening direction — the sign of the constant decides which way the parabola opens. For x^2 = 4Ay with A < 0, the parabola opens downward, the axis is x = 0, and the latus rectum is the line y = A, that is 4y - 4A = 0. e.g. For x^2 = -8y, the latus rectum is y = -2.
• directrix from a parameter — for y^2 = 4bx the directrix is x = -b, so a negative b places the parabola on the left and the directrix on the right. When b is fixed by a quadratic condition, factor the condition and keep the root that satisfies the sign restriction.
• area of a parabolic segment — the region between y^2 = 4kx and its latus rectum has area \frac{2}{3} \times (\text{base}) \times (\text{height}), with base 4k and height k, which gives \frac{8}{3}k^2.
• two parabolas from one latus rectum — if the endpoints of a latus rectum are fixed, the focus is their midpoint, and the vertex may lie at distance a from the focus on either side. This gives two parabolas with the same focus whose axes are the same line but which open in opposite directions.
• latus rectum of an ellipse — for \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, the latus rectum through the focus (ae, 0) is the chord x = ae, with endpoints at y = \pm \frac{b^2}{a} and length \frac{2b^2}{a}. Every point of this chord, including the focus, satisfies x = ae.

Circle equation radius

The general equation of a circle is x^2 + y^2 + 2gx + 2fy + c = 0. Its centre is (-g, -f) and its radius is \sqrt{g^2 + f^2 - c}. Before reading off the centre and radius, divide through by the coefficient of x^2 so that the coefficients of x^2 and y^2 are both 1. Completing the square gives (x + g)^2 + (y + f)^2 = g^2 + f^2 - c, and a real circle requires the right-hand side to be positive.

• radius from the normalised form — with g = -3, f = 2 and c = -3, the centre is (3, -2) and the radius is \sqrt{9 + 4 + 3} = 4. The radius is always the square root of g^2 + f^2 - c, not of c alone.
• normalising the coefficients — if the x^2 and y^2 terms have a common coefficient k, divide the whole equation by k before comparing with the standard form. Otherwise the centre and radius are read off incorrectly.
• circle through three points — substituting three points into x^2 + y^2 + 2gx + 2fy + c = 0 gives three linear equations in g, f and c. Solving them gives the centre (-g, -f), and the radius is the distance from the centre to any of the three points.
• radius as a distance — the radius equals the distance from the centre to any point on the circle, so a point at distance r from the centre fixes which range r belongs to.
• circle touching axes and a line — a circle touching both positive coordinate axes in the first quadrant has its centre at (r, r) and radius r. If it also touches a horizontal line y = k, the distance from the centre to that line equals the radius, so |r - k| = r.
• diameter — the diameter of a circle is 2r, not r, and not r multiplied by any other factor.
• centre and radius as parameters — if the centre is (a, b) and the radius is c, read a = -g, b = -f and c^2 = g^2 + f^2 - c_0 from the equation before combining them, for example in the form a^2 + b^2 + c^2.

Parallelogram vertices

In a parallelogram ABCD, the diagonals bisect each other, so the midpoint of AC equals the midpoint of BD. From this, if three consecutive vertices A, B and C are known, the fourth vertex is D = A + C - B, which is equivalent to \vec{AD} = \vec{BC}.

• fourth vertex — with B lying between A and C, the missing vertex is D = A + C - B. Check the answer by confirming that the two diagonals have the same midpoint.
• equation of a diagonal — a diagonal is the line through two opposite vertices, so use the two-point form with those vertices. Verify by substituting both vertices into the final equation.
• area by cross product — if adjacent sides are \vec{u} = (u_1, u_2) and \vec{v} = (v_1, v_2), the area of the parallelogram is |u_1v_2 - u_2v_1|. A diagonal divides the parallelogram into two triangles of equal area, so the area is also twice the area of the triangle formed by three consecutive vertices.
• diagonal through the origin — when the vertices are the pairwise intersections of given lines, each vertex comes from solving a pair of equations. A diagonal through the origin joins the origin to the vertex opposite it, and if that vertex is (x_0, y_0), its equation is y_0 x - x_0 y = 0.
• line through two points on the axes — the line joining the x-intercept (p, 0) and the y-intercept (0, q) has intercept form \frac{x}{p} + \frac{y}{q} = 1. Clearing the fractions gives the general form.

Eccentricity of conics

Eccentricity e measures how far a conic departs from a circle. The standard values are: circle e = 0, ellipse 0 < e < 1, parabola e = 1, and hyperbola e > 1. For the ellipse \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 with a > b, which has its major axis along the x-axis, b^2 = a^2(1 - e^2), so e = \sqrt{1 - \frac{b^2}{a^2}}. If the larger denominator is under y, the major axis lies on the y-axis and the roles of a and b are interchanged. For the hyperbola \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, c^2 = a^2 + b^2 and e = \sqrt{1 + \frac{b^2}{a^2}}.

• eccentricity from parametric form — the identity \sec^2\theta - \tan^2\theta = 1 eliminates \theta. The variable with the \sec term belongs to the transverse axis, so its denominator is a^2 in the formula for e.
• the standard ranges — the values e = 0 for a circle, e = 1 for a parabola and e < 1 for an ellipse are standard facts, and each one holds without exception.
• ellipse with the major axis on the y-axis — solve for the two unknowns a^2 and b^2 from the two given points, then take the larger of them as the denominator under y. Then e^2 = 1 - \frac{\text{smaller}}{\text{larger}}.
• foci of a family of conics — for \frac{x^2}{\cos^2\theta} - \frac{y^2}{\sin^2\theta} = 1, c^2 = \cos^2\theta + \sin^2\theta = 1, so the foci are (\pm 1, 0) for every \theta. The eccentricity is e = \frac{c}{a} = \sec\theta when \cos\theta > 0, and the distance between the foci is 2c.
• focal radii at the end of the minor axis — if the two focal radii from an end of the minor axis are perpendicular, then b = ae. Combined with b^2 = a^2(1 - e^2), this gives e^2 = \frac{1}{2}, so e = \frac{1}{\sqrt{2}}.

Family of lines through fixed point

A line whose equation contains a parameter, such as m, is a member of a family of lines. The family is concurrent when every member passes through one fixed point (x_0, y_0). To find this point, rearrange the equation into the form P(x, y) + mQ(x, y) = 0 and solve P = 0 and Q = 0 simultaneously. The lines through the intersection of two given lines L_1 = 0 and L_2 = 0 take the form L_1 + \lambda L_2 = 0.

• fixed point from a linear constraint — when the parameters satisfy a linear relation, such as A, B, C in arithmetic progression giving 2B = A + C, substitute it to obtain an expression of the form A(x + y) + C(y + 1) = 0. Both coefficient expressions must vanish for the point to lie on every line, which gives the fixed point.
• intersection of two lines — the point where two lines meet is found by solving their equations simultaneously. A line through that point with equal intercepts on the axes has the form x + y = a when the intercepts have the same sign, and x - y = a when they are equal in magnitude with opposite signs. The sum of the absolute values of the intercepts is then computed from the intercept form.
• locus of the midpoint of intercepts — for a line \frac{x}{a} + \frac{y}{b} = 1 meeting the axes at (a, 0) and (0, b), the midpoint of the segment is \left(\frac{a}{2}, \frac{b}{2}\right). Express a and b in terms of the midpoint coordinates, substitute into the fixed-point condition, and eliminate the parameters to obtain the locus.
• concurrent lines — an equation that is linear in m, such as 4y = m(x - 1) + 2, can be checked by setting the coefficient of m and the constant term to zero. Both conditions hold at the point (1, \frac{1}{2}), so every line of the family passes through it and the lines are concurrent.

Perpendicular lines condition

Two non-vertical lines with slopes m_1 and m_2 are perpendicular if and only if m_1 m_2 = -1. For lines in general form, a_1x + b_1y + c_1 = 0 and a_2x + b_2y + c_2 = 0, the perpendicularity condition is a_1a_2 + b_1b_2 = 0, which avoids computing slopes. The slope of ax + by + c = 0 is -\frac{a}{b}. A line perpendicular to px + qy = r has the form qx - py + k = 0, and the constant k is found by substituting a given point.

• product of slopes — write each line in the form y = mx + c and set the product of the two slopes equal to -1. A line with a slope of -p and another with a slope of q are perpendicular exactly when pq = 1.
• altitude of a triangle — the altitude from a vertex to the opposite side is perpendicular to that side and passes through the vertex. Its slope is the negative reciprocal of the side's slope, and the altitude is then found from the point-slope form.
• normal vector and perpendicular line — the normal vector of px + qy = r is (p, q). A line perpendicular to it has a normal vector parallel to the direction of the original line, which gives the equation qx - py + k = 0. The constant k is determined by substituting the point the line must pass through, and the trigonometric identity \sin^2\theta + \cos^2\theta = 1 often simplifies this step.

Collinearity of points

Three points P_1(x_1, y_1), P_2(x_2, y_2) and P_3(x_3, y_3) are collinear if and only if the triangle they form has zero area, that is x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) = 0. Equivalently, the slope of P_1P_2 equals the slope of P_2P_3, provided the denominators are non-zero. The collinearity condition may hold for every value of a parameter, in which case the points always lie on one line. Coordinates may be negative, so a family of points is not confined to a single quadrant unless its coordinates are forced to be positive.

• determinant test — evaluate the determinant of the coordinate matrix with a row of 1s. If it is identically zero in the parameters, the points are collinear for all values of those parameters. The quadrant of the points depends on the signs of their coordinates, which the parameters can change.
• equal slopes with parameters — equate the slopes between pairs of points and simplify. Factor the result, and when the problem requires two parameters to be distinct, discard the factor that forces them to be equal. For example, (p - q)(pq - 1) = 0 with p \ne q leaves pq = 1.
• points on a line through the origin — if every point satisfies y = kx, the points are collinear. For instance, the points (-a, -b), (0, 0) and (a, b) satisfy y = \frac{b}{a}x, and so does (a^2, ab). Four collinear points cannot be the vertices of a square or parallelogram, and they cannot lie on a circle, because a line meets a circle in at most two points.

Angle between two lines

The angle between two lines is the angle between their direction vectors, and the acute angle is the one usually asked for. If a line has inclination \alpha with the positive x-axis, its slope is m=\tan\alpha. A vertical line has inclination 90^\circ and no finite slope, so the angle between a vertical line and a line of inclination \alpha is 90^\circ-\alpha. For two non-vertical lines with slopes m_1 and m_2, the acute angle \theta satisfies $\tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right|.$ The lines are parallel when m_1=m_2, which gives \theta=0, and perpendicular when m_1m_2=-1, which makes the denominator vanish. The condition \tan\theta=1 gives \theta=45^\circ and \sin\theta=\frac{1}{\sqrt{2}}. In an isosceles triangle with apex angle A, each equal base angle is \theta=\frac{\pi-A}{2}, so \cot\theta=\tan\frac{A}{2}=\frac{\sin A}{1+\cos A}. Example: for slopes 2 and \frac{1}{3}, \tan\theta=\left|\frac{2-\frac{1}{3}}{1+\frac{2}{3}}\right|=1, so the acute angle is 45^\circ.

Equilateral triangle vertices

An equilateral triangle with side s has all sides equal and all angles 60^\circ. Its height is \frac{\sqrt{3}}{2}s and its area is \frac{\sqrt{3}}{4}s^2. If two vertices A and B are known, the third vertex lies on the perpendicular bisector of AB, at distance \frac{\sqrt{3}}{2}|AB| from the midpoint of AB. There are two such points, one on each side of AB. Because \sqrt{3} often appears in these coordinates, the third vertex can have irrational coordinates and the area can be irrational even when the given vertices are rational. For example, for A(0,0) and B(2,0) the third vertex is (1,\pm\sqrt{3}), and the area is \sqrt{3}.

When one vertex of the equilateral triangle is at the vertex of a parabola, the other two vertices can be written as (\pm t,\,y(t)) on the parabola. The condition that all sides are equal then gives an equation in t. Equating the side from the vertex to the side between the other two vertices determines t, and the side length is then found directly.

Diagonals of a square

The diagonals of a square are equal in length, bisect each other at right angles, and meet at the centre of the square. The centre is the midpoint of each diagonal. If the centre is O and one vertex is A, the opposite vertex is C=2O-A in vector form. A diagonal of length d gives a side of \frac{d}{\sqrt{2}} and an area of \frac{d^2}{2}. Two diagonals through the centre are perpendicular, so their slopes m_1 and m_2 satisfy m_1m_2=-1. For example, if the centre is (1,1) and one vertex is (3,0), the opposite vertex is (-1,2), so the diagonal length is \sqrt{16+4}=2\sqrt{5} and the area is 10 square units.

Linear inequalities region

A linear inequality such as ax+by+c<0 represents a half-plane bounded by the line ax+by+c=0. The line itself is included for \le or \ge and excluded for strict < or >. To find which side is the solution, test a convenient point not on the line, often the origin. The common region of a system is the intersection of all the half-planes. A candidate point is in the region only if it satisfies every inequality at once, so the test is done for all of them. For example, the point (1,1) satisfies x+2y\le 4 because 1+2=3\le 4.

An absolute value inequality describes a rectangle. The inequality |x|\le a means -a\le x\le a, and |y|\le b means -b\le y\le b. The region is therefore a rectangle with side lengths 2a and 2b, and its area is 4ab. This is different from taking a and b as side lengths directly.

Parabola equation from focus and directrix

A parabola is the locus of points equidistant from a fixed point, the focus, and a fixed line, the directrix. The vertex lies midway between the focus and the directrix, and the axis passes through the focus perpendicular to the directrix. The standard forms with vertex at the origin are

• y^2=4ax, focus (a,0), directrix x=-a, opens right
• y^2=-4ax, focus (-a,0), directrix x=a, opens left
• x^2=4ay, focus (0,a), directrix y=-a, opens up
• x^2=-4ay, focus (0,-a), directrix y=a, opens down

The latus rectum has length 4a. Example: focus (0,2) and directrix y=-2 give a=2 and x^2=8y.

Foci of ellipse

For the ellipse \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 with a>b>0, the major axis lies along the x-axis. The foci are at (\pm c,0), where c^2=a^2-b^2, and the distance between the foci is 2c. If b>a, the major axis lies along the y-axis and the foci are at (0,\pm c) with c^2=b^2-a^2. Before using the formula, divide the equation so that the right side is 1, because a^2 and b^2 are read from the denominators. For any point P on the ellipse, the sum of distances to the two foci equals the length of the major axis, PF_1+PF_2=2a. Example: for \frac{x^2}{25}+\frac{y^2}{9}=1, c^2=16, so the foci are (\pm4,0) and the focal distance is 8.