Parabola — standard forms (4 orientations), focus, directrix, latus rectum, axis: Exercises

The single sentence behind all of it (from the parent): $distance (P, focus) = distance (P, directrix) .$ Everything below is just that sentence, read carefully.

Level 1 — Recognition

Here you only decide which of the four standard forms you are looking at, and read off the meaning of the number in it. Recall the map from the parent note:

Exercise 1.1

For $y^{2} = 20 x$ , state $a$ , the direction it opens, the focus, and the directrix.

Recall Solution 1.1

WHAT form is this? $y^{2}$ present, sign positive $\Rightarrow$ matches $y^{2} = 4 a x$ , opens right. Find $a$ . The coefficient of $x$ is the whole thing $4 a$ : $4 a = 20 ⟹ a = 5.$ Read off using the $y^{2} = 4 a x$ row:

Focus $(a, 0) = (5, 0)$ .
Directrix $x = - a \Rightarrow x = - 5$ .
Axis $y = 0$ ; latus rectum $= 4 a = 20$ .

Exercise 1.2

For $x^{2} = - 16 y$ , state $a$ , the direction, the focus, and the directrix.

Recall Solution 1.2

WHAT form? $x^{2}$ present $\Rightarrow$ vertical axis. Sign negative $\Rightarrow$ opens down, matching $x^{2} = - 4 a y$ . Find $a$ . $4 a = 16 ⟹ a = 4$ . Read off the $x^{2} = - 4 a y$ row:

Focus $(0, - a) = (0, - 4)$ .
Directrix $y = a \Rightarrow y = 4$ .
Axis $x = 0$ ; latus rectum $= 4 a = 16$ .

Level 2 — Application

Now you compute several quantities together and, in one problem, run the definition itself.

Exercise 2.1

Find the endpoints of the latus rectum of $y^{2} = 12 x$ , and its length.

Recall Solution 2.1

Find $a$ : $4 a = 12 \Rightarrow a = 3$ . Opens right; focus $(3, 0)$ . WHY substitute $x = a$ ? The latus rectum is the chord through the focus, parallel to the directrix. The directrix is vertical, so this chord is the vertical line through the focus: $x = a = 3$ . Put $x = 3$ into $y^{2} = 12 x$ : $y^{2} = 12 \cdot 3 = 36 ⟹ y = \pm 6.$ Endpoints: $(3, 6)$ and $(3, - 6)$ . Length: $6 - (- 6) = 12$ , which equals $4 a = 12$ . ✓ (Look at figure below — the red chord.)

Figure — Parabola — standard forms (4 orientations), focus, directrix, latus rectum, axis

Exercise 2.2

A parabola has focus $(- 3, 0)$ and directrix $x = 3$ . Use the definition directly (do not quote a standard form) to find its equation.

Recall Solution 2.2

WHAT is the definition? For a point $P = (x, y)$ on the curve, $distance (P, focus) = distance (P, directrix) .$

Distance to focus $(- 3, 0)$ : $(x + 3)^{2} + y^{2}$ (this is the distance formula).
Distance to the vertical line $x = 3$ : the horizontal gap $∣ x - 3∣$ . Set equal and square (squaring is safe — both sides are $\geq 0$ , so no solutions are lost or gained): $(x + 3)^{2} + y^{2} = (x - 3)^{2} .$ Expand: $x^{2} + 6 x + 9 + y^{2} = x^{2} - 6 x + 9$ . Cancel $x^{2}$ and $9$ : $y^{2} = - 12 x .$ Sanity check: focus is on the negative- $x$ side, so it should open left — and $y^{2} = - 12 x$ (negative sign, $y^{2}$ present) does open left. ✓ Here $4 a = 12 \Rightarrow a = 3$ , matching focus $(- a, 0) = (- 3, 0)$ .

Level 3 — Analysis

Here the parabola is not centred at the origin, or the form is disguised. You must think, not just match.

Exercise 3.1

Find the vertex, focus, and directrix of $y^{2} - 8 y - 4 x + 20 = 0.$

Recall Solution 3.1

WHY complete the square? The standard forms all have a single squared variable and a linear other variable. This equation has $y^{2}$ and $y$ together — so we bundle them into one perfect square to expose a shifted standard form. Group the $y$ terms: $y^{2} - 8 y = (y - 4)^{2} - 16.$ Substitute back: $(y - 4)^{2} - 16 - 4 x + 20 = 0 ⟹ (y - 4)^{2} = 4 x - 4 = 4 (x - 1) .$ WHAT does this say? Let $Y = y - 4$ and $X = x - 1$ . Then $Y^{2} = 4 X$ , the right-opening standard form with $4 a = 4 \Rightarrow a = 1$ , in shifted coordinates whose origin (the vertex) is at $(x, y) = (1, 4)$ . Translate everything back (add the shift $(1, 4)$ ):

Vertex: $(1, 4)$ .
Focus: $X = a, Y = 0 \Rightarrow (x, y) = (1 + 1, 4 + 0) = (2, 4)$ .
Directrix: $X = - a \Rightarrow x - 1 = - 1 \Rightarrow x = 0$ .
Axis: $Y = 0 \Rightarrow y = 4$ ; latus rectum $= 4 a = 4$ .

Exercise 3.2

For which value(s) of $k$ does $x^{2} = k y$ have its focus at $(0, - 2)$ ? Give the directrix too.

Recall Solution 3.2

Analyse the sign. Focus at $(0, - 2)$ is below the vertex $(0, 0)$ , so the parabola opens down, matching $x^{2} = - 4 a y$ with $a > 0$ . For the downward form the focus is $(0, - a)$ , so: $- a = - 2 ⟹ a = 2.$ Then $x^{2} = - 4 a y = - 8 y$ , i.e. $k = - 8$ . Directrix of the downward form is $y = a = 2$ . Note the trap avoided: you cannot just write $x^{2} = k y$ and set $k = - 2$ — the coefficient is $- 4 a$ , not $- a$ .

Level 4 — Synthesis

Build the equation from geometric data, sometimes with the vertex moved.

Exercise 4.1

Find the equation of the parabola with vertex $(0, 0)$ that passes through $(2, - 6)$ and is symmetric about the $x$ -axis.

Recall Solution 4.1

Choose the form. Symmetric about the $x$ -axis $\Rightarrow$ axis is $y = 0 \Rightarrow$ the form is $y^{2} = 4 a x$ (or $y^{2} = - 4 a x$ ). We don't yet know the sign, so keep it as $y^{2} = c x$ and let the point decide $c$ . Use the point $(2, - 6)$ : $(- 6)^{2} = c \cdot 2 ⟹ 36 = 2 c ⟹ c = 18.$ So $y^{2} = 18 x$ . Since $c = 18 > 0$ , it opens right — consistent with the point having positive $x$ . Extract $a$ : $4 a = 18 \Rightarrow a = 4.5$ ; focus $(4.5, 0)$ , directrix $x = - 4.5$ , latus rectum $18$ .

Exercise 4.2

A parabola has vertex $(2, 3)$ , opens upward, and its latus rectum has length $8$ . Find its equation, focus, and directrix.

Recall Solution 4.2

Choose the shifted form. Opens up with vertex $(h, k) = (2, 3)$ : $(x - h)^{2} = 4 a (y - k) ⟹ (x - 2)^{2} = 4 a (y - 3) .$ Use the latus rectum. Length $= 4 a = 8 \Rightarrow a = 2$ . So $4 a = 8$ : $(x - 2)^{2} = 8 (y - 3) .$ Focus and directrix — measure $a = 2$ from the vertex along the axis of symmetry $x = 2$ :

Opens up, so focus is $a$ above the vertex: $(2, 3 + 2) = (2, 5)$ .
Directrix is $a$ below: $y = 3 - 2 = 1$ . Check: vertex $(2, 3)$ is midway between focus $y = 5$ and directrix $y = 1$ ; midpoint $y = 3$ . ✓

Level 5 — Mastery

Combine the parabola with a second concept: a chord condition, and a real trajectory.

Exercise 5.1

The parabola $y^{2} = 8 x$ carries a point $P$ whose distance from the focus equals $6$ . Find the coordinates of all such $P$ .

Recall Solution 5.1

WHY the definition beats the distance formula here. By the very definition, distance to focus $=$ distance to directrix. That second distance is just a horizontal gap — far easier than a square root. Set up. $4 a = 8 \Rightarrow a = 2$ . Directrix is $x = - a = - 2$ . For a point $(x, y)$ on the curve (so $x \geq 0$ ), the distance to the directrix is $x - (- 2) = x + 2$ . Distance to focus $= 6$ means: $x + 2 = 6 ⟹ x = 4.$ Find $y$ from the parabola: $y^{2} = 8 \cdot 4 = 32 \Rightarrow y = \pm 32 = \pm 42$ . Both points: $(4, 42)$ and $(4, - 42)$ . (See figure — both dots are the same distance from the focus.)

Exercise 5.2

A ball is thrown and its path is the parabola $y = x - \frac{1}{20} x^{2}$ (with $x, y$ in metres, ground at $y = 0$ ). Find the maximum height, and rewrite the path in the vertex standard form $(x - h)^{2} = - 4 a (y - k)$ to read off $a$ and hence the latus rectum. (Connects to Projectile Motion.)

Recall Solution 5.2

Rewrite by completing the square. Factor out the $x^{2}$ coefficient: $y = - \frac{1}{20} (x^{2} - 20 x) = - \frac{1}{20} ((x - 10)^{2} - 100) = - \frac{1}{20} (x - 10)^{2} + 5.$ Read the vertex. Vertex $(h, k) = (10, 5)$ , so the maximum height is $5$ m (at horizontal distance $x = 10$ m). Put in standard form. From $y - 5 = - \frac{1}{20} (x - 10)^{2}$ multiply both sides by $- 20$ : $(x - 10)^{2} = - 20 (y - 5) .$ This is $(x - h)^{2} = - 4 a (y - k)$ (opens down, as a thrown ball must). Match coefficients: $4 a = 20 ⟹ a = 5, latus rectum = 4 a = 20 m .$ Interpretation: the ball's arch is a downward parabola with vertex at its peak — exactly why projectile trajectories are parabolic.

Wrap-up recall

Recall One-line answers (forecast, then reveal)

Ex 1.1 focus / directrix of $y^{2} = 20 x$ ::: $(5, 0)$ / $x = - 5$ Ex 1.2 focus / directrix of $x^{2} = - 16 y$ ::: $(0, - 4)$ / $y = 4$ Ex 2.1 latus rectum endpoints of $y^{2} = 12 x$ ::: $(3, 6)$ and $(3, - 6)$ , length $12$ Ex 2.2 equation, focus $(- 3, 0)$ , directrix $x = 3$ ::: $y^{2} = - 12 x$ Ex 3.1 vertex / focus / directrix of $y^{2} - 8 y - 4 x + 20 = 0$ ::: $(1, 4)$ / $(2, 4)$ / $x = 0$ Ex 3.2 value of $k$ so $x^{2} = k y$ has focus $(0, - 2)$ ::: $k = - 8$ , directrix $y = 2$ Ex 4.1 parabola through $(2, - 6)$ , vertex origin, $x$ -axis symmetric ::: $y^{2} = 18 x$ Ex 4.2 vertex $(2, 3)$ , opens up, LR $= 8$ ::: $(x - 2)^{2} = 8 (y - 3)$ , focus $(2, 5)$ , directrix $y = 1$ Ex 5.1 points on $y^{2} = 8 x$ at focal distance $6$ ::: $(4, \pm 42)$ Ex 5.2 max height / latus rectum of $y = x - \frac{1}{20} x^{2}$ ::: $5$ m / $20$ m

Connections

Parent topic note
Conic Sections — Overview
Coordinate Geometry — Distance Formula (used in Ex 2.2)
Tangents and Normals to a Parabola (next after these exercises)
Projectile Motion (Ex 5.2)
Ellipse — standard forms · Hyperbola — standard forms · Eccentricity and focus-directrix definition (compare the other conics)