The translated parabola

Consider a translated parabola (Figure 1.6). The distance from a moving point on the parabola to the focus is equal to the distance from that point to the directrix (FP = PM). VF = VD = a. The coordinates of F are (h, a + k), and the equation of the directrix is x = h - a.

Translated parabolas

If $P(x, y)$ is any moving point on the parabola, then $FP = PM$. The coordinates of $M$ and $F$ are $(h - a, y)$ and $(h + a, y)$, respectively.

Using the distance formula and the definition of a parabola:

$\sqrt{(x - (h + a))^2 + (y - k)^2} = |x - (h - a)|$

Squaring both sides and simplifying leads to:

$(y - k)^2 = 4a(x - h)$ (Equation 1.1)

Equation 1.1 represents a translated parabola opening to the right, where the directrix is parallel to the y-axis.

If the vertex is translated $h$ units horizontally and $k$ units vertically, the new vertex is at $(h, k)$. Both the focus and directrix are also moved $h$ units horizontally and $k$ units vertically.

Other forms of translated parabolas:

• $(y - k)^2 = -4a(x - h)$ represents a translated parabola opening to the left.
• $(x - h)^2 = 4b(y - k)$ represents a translated parabola opening upwards.
• $(x - h)^2 = -4b(y - k)$ represents a translated parabola opening downwards.

Example 1

Find the equation of the parabola with focus at $(-4, 0)$, directrix $x = -8$, and vertex at $(-6, 0)$.

Using $FP = PM$ (or $FC=CD$ as in your example):

$\sqrt{(x + 4)^2 + y^2} = |x + 8|$

Squaring both sides and simplifying:

$(x + 4)^2 + y^2 = (x + 8)^2$

$x^2 + 8x + 16 + y^2 = x^2 + 16x + 64$

$y^2 = 8x + 48$

$y^2 = 8(x + 6)$

Example 2

Given the parabola $x^2 + 4x + 4y + 16 = 0$, find:

1. The coordinates of the vertex.
2. The coordinates of the focus.
3. The equation of the directrix.

Rewrite the equation in standard form by completing the square:

$(x + 2)^2 = -4(y + 3)$

Here, $h = -2$, $k = -3$, and $b = 1$.

(a) Vertex: $V(h, k) = V(-2, -3)$

(b) Focus: $F(h, k - b) = F(-2, -3 - 1) = F(-2, -4)$

(c) Directrix: $y = k + b = y = -3 + 1 = y = -2$

Example 3

A bridge rope forms a parabola. The towers are 24 meters high and 144 meters apart. The lowest point of the rope is 4 meters above the road. Find:

1. The equation of the parabola (road as x-axis, parabola's axis as y-axis).
2. The height of the rope 36 meters from the center of the road.

Solution:

a. The vertex is at $(0, 4)$. The towers are at $(-72, 24)$ and $(72, 24)$. The equation will be of the form $x^2 = 4b(y - 4)$.

Using the point $(72, 24)$:

$72^2 = 4b(24 - 4)$

$5184 = 80b$

$b = \frac{5184}{80} = 64.8$

So, the equation is $x^2 = 4(64.8)(y - 4)$ or $x^2 = 259.2(y - 4)$

b. To find the height 36 meters from the center, let $x = 36$:

$36^2 = 259.2(y - 4)$

$1296 = 259.2(y - 4)$

$5 = y - 4$

$y = 9$

The height of the rope 36 meters from the center is 9 meters.

A tangent is a line that touches the parabola at only one point. A normal is a line perpendicular to the tangent at the point of tangency.

Let's consider the parabola $y^2 = 4ax$ and a point of tangency $P(x_1, y_1)$. The gradient ($m$) of the tangent is the first derivative of the parabola's equation:

$\frac{dy}{dx} = \frac{2a}{y}$

At point $P(x_1, y_1)$, the gradient of the tangent is:

$m = \frac{2a}{y_1}$

Equation of the tangent

The equation of the tangent at $P(x_1, y_1)$ is given by:

$y - y_1 = m(x - x_1)$

Substituting $m = \frac{2a}{y_1}$:

$y - y_1 = \frac{2a}{y_1}(x - x_1)$

Cross-multiplying and simplifying (using $y_1^2 = 4ax_1$):

$yy_1 - y_1^2 = 2ax - 2ax_1$

$yy_1 - 4ax_1 = 2ax - 2ax_1$

$yy_1 = 2a(x + x_1)$

Equation of the normal

The normal is perpendicular to the tangent. The product of the gradients of perpendicular lines is -1 ($m_1m_2 = -1$). If $m_1$ is the gradient of the tangent, then the gradient of the normal ($m_2$) is:

$m_2 = -\frac{y_1}{2a}$

The equation of the normal at $P(x_1, y_1)$ is:

$y - y_1 = m_2(x - x_1)$

$y - y_1 = -\frac{y_1}{2a}(x - x_1)$

Simplifying:

$2ay - 2ay_1 = -y_1x + x_1y_1$

$y_1x + 2ay - x_1y_1 - 2ay_1 = 0$

Example 1

Find the equations of the tangent and normal at $(3, 4)$ on the parabola $y^2 = 4ax$.

Substituting $(3,4)$ into $y^2=4ax$ gives $16=12a$, so $a=\frac{4}{3}$.

$\frac{dy}{dx} = \frac{2a}{y} = \frac{2(\frac{4}{3})}{4} = \frac{2}{3}$

Tangent: $y - 4 = \frac{2}{3}(x - 3) \Rightarrow 3y - 12 = 2x - 6 \Rightarrow 2x - 3y + 6 = 0$

Normal: Gradient is $-\frac{3}{2}$. $y - 4 = -\frac{3}{2}(x - 3) \Rightarrow 2y - 8 = -3x + 9 \Rightarrow 3x + 2y - 17 = 0$

Example 2

Find the equations of the tangent and normal at the vertex of the parabola $x^2 + 4x - 8y - 4 = 0$.

Rewrite: $(x + 2)^2 = 8(y + 1)$. Vertex is $(-2, -1)$.

Differentiating: $2(x+2) = 8\frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{x+2}{4}$

At $x = -2$, $\frac{dy}{dx} = 0$ (horizontal tangent).

Tangent: $y = -1$

Normal: $x = -2$ (vertical line)

Example 3

Show that the condition for $y = mx + c$ to be a tangent to $y^2 = 4ax$ is $c = \frac{a}{m}$. Find the point of tangency.

Substitute $y = mx + c$ into $y^2 = 4ax$: $(mx + c)^2 = 4ax$

$m^2x^2 + 2mcx + c^2 - 4ax = 0$

$m^2x^2 + (2mc - 4a)x + c^2 = 0$

For tangency, the discriminant is zero: $(2mc - 4a)^2 - 4m^2c^2 = 0$

$4m^2c^2 - 16amc + 16a^2 - 4m^2c^2 = 0$

$-16amc + 16a^2 = 0 \Rightarrow c = \frac{a}{m}$

Substituting $c = \frac{a}{m}$ back into the quadratic equation and solving for $x$ gives $x=\frac{a}{m^2}$, then substituting $x$ into $y=mx+c$ gives $y=2a/m$. Thus point of tangency $\left(\frac{a}{m^2},\frac{2a}{m}\right)$

Example 4

Find $n$ so that $y = x + n$ is tangent to $y^2 = 4x$.

Here $m=1$, $c=n$ and $a=1$. Using $c = \frac{a}{m}$, we find $n=1$

Alternative approach: Substitute $y=x+n$ into $y^2 = 4x$: $(x+n)^2 = 4x \Rightarrow x^2 + 2nx + n^2 - 4x = 0$

$x^2 + (2n-4)x + n^2 = 0$

For tangency, discriminant = 0: $(2n-4)^2 - 4n^2 = 0 \Rightarrow 4n^2 - 16n + 16 - 4n^2 = 0 \Rightarrow -16n = -16 \Rightarrow n=1$

Example 5

The line $y = mx - 2$ touches $y^2 = 2x$. Determine:

1. Possible values of $m$.
2. Points of contact.

Substituting $y = mx - 2$ into $y^2 = 2x$: $(mx - 2)^2 = 2x \Rightarrow m^2x^2 - 4mx + 4 = 2x \Rightarrow m^2x^2 - (4m+2)x + 4 = 0$.

Discriminant=0: $(4m+2)^2 - 16m^2 = 0 \Rightarrow 16m^2 + 16m + 4 - 16m^2 = 0 \Rightarrow 16m = -4 \Rightarrow m = -\frac{1}{4}$

Substituting $m=-\frac{1}{4}$ back into the quadratic equation gives $x=4$, then substituting $x=4$ into $y=mx-2$ gives $y=-3$. Point of contact is $(4,-3)$

Length of latus rectum

The latus rectum is the line segment LL′ passing through the focus and perpendicular to the axis of symmetry. The x-coordinates of L and L′ are 'a' (the x-coordinate of the focus F(a, 0)).

If $L$ has coordinates $(a, b)$, and $L$ lies on the parabola $y^2 = 4ax$, then:

$b^2 = 4a(a)$

$b^2 = 4a^2$

$b = \pm 2a$

The coordinates of $L$ and $L'$ are $(a, 2a)$ and $(a, -2a)$, respectively.

The length of the latus rectum $LL'$ is:

$LL' = 2a - (-2a) = 4a$

Therefore, the length of the latus rectum is $4a$ units.

Example 1

Find the length of the latus rectum of $y^2 = 24x$.

Comparing $y^2 = 24x$ with $y^2 = 4ax$ gives $4a = 24$, so $a = 6$.

Length of latus rectum $= 4a = 4 * 6 = 24$ units.

Example 2

Find the length of the latus rectum of $y^2 - 4y - 16x + 68 = 0$.

Completing the square: $(y - 2)^2 = 16(x - 4)$.

Comparing with $(y - k)^2 = 4a(x - h)$ gives $4a = 16$.

Length of latus rectum $= 16$ units.

Parametric equations of a parabola

Parametric equations express coordinates in terms of a parameter (e.g., $t$). For the parabola $y^2 = 4ax$, the parametric equations are:

$x = at^2$

$y = 2at$

The parametric coordinate point is $(at^2, 2at)$.

Example 1

Find the parametric equations of $y^2 = 8x$.

Comparing with $y^2 = 4ax$ gives $4a = 8$, so $a = 2$.

$x = 2t^2$

$y = 4t$

Example 2

Express the following parabolas in parametric form:

(a) $x^2 = -5y$

(b) $(x - 2)^2 = -8(y - 3)$

(a) Comparing with $x^2 = -4ay$ gives $-4a = -5$, so $a = \frac{5}{4}$.

$x = \frac{5}{4}t^2$

$y = -\frac{5}{2}t$

(b) Comparing with $(x - h)^2 = -4a(y - k)$ gives $h=2$, $k=3$ and $4a=8$ so $a=2$.

$x-2=2t^2$ so $x=2t^2+2$

$y-3=-4t$ so $y=-4t+3$

Example 3

Prove that $x = t^2+3$ and $y = \frac{1}{2}(3t + 1)$ are parametric equations of a parabola.

Solve for $t$ in the $y$ equation: $2y-1=3t$ so $t=\frac{2y-1}{3}$

Substituting $t$ into the $x$ equation gives $x=\left(\frac{2y-1}{3}\right)^2+3$

Simplify: $x=\frac{4y^2-4y+1}{9}+3 \Rightarrow 9x=4y^2-4y+1+27 \Rightarrow 9x=4y^2-4y+28$. This is a parabola.