# Find the latus rectum of the Parabola

Let $y=3x-8$ be the equation of tangent at the point $(7,13)$ lying on a parabola, whose focus is at $(-1,-1)$. Evaluate the length of the latus rectum of the parabola.

I got this question in my weekly test. I tried to assume the general equation of the parabola and solve the system of equations to calculate the coefficients with the help of these given conditions. But this way it becomes very lengthy and tedious. Can anyone provide an elegant solution? Thanks.

• is the axis of the parabola meant to be vertical? – David Quinn Apr 18 '16 at 15:37
• No. Everything given in the question is sufficient according to my teacher. It is a general parabola. – Henry Apr 18 '16 at 16:10

Let $P(7,13),F(-1,-1)$. Also, let $T$ be the intersection point of the line $y=3x-8$ with the axis of symmetry. Let $V$ be the vertex, and let $K$ be the point on the axis of symmetry such that $PK$ is perpendicular to the axis.

We use the followings (for the proof, see the end of this answer) :

(1) $PF=TF$

(2) $VT=VK$

(3) $\text{(the length of the latus rectum)}=4\times FV$

First of all, setting $T$ as $(t,3t-8)$ where $t\not= 7$ and using $(1)$ give $$(-1-7)^2+(-1-13)^2=(-1-t)^2+(-1-3t+8)^2\quad\Rightarrow\quad t=-3\quad\Rightarrow\quad T(-3,-17)$$

Hence, the axis of symmetry is the line $TF$ : $y=8x+7$. So, the line $PK$ is $y-13=(-1/8)(x-7)$, i.e. $y=-x/8+111/8$ from which $K(11/13,179/13)$ follows.

From $(2)$, since $V$ is the midpoint of the line segment $TK$, we have $V(-14/13,-21/13).$

Finally, using $(3)$, we get that the answer is $\color{red}{4\sqrt{5/13}}$.

Proof for $(1)(2)(3)$ :

We may suppose that the equation of a parabola is $y^2=4px$ where $p\gt 0$.

$\qquad\qquad\qquad$ We consider the tangent line at $A(a,b)$ where $b^2=4pa$ with $b\gt 0$. Let $B$ be the intersection point of the tangent line with $x$ axis which is the axis of symmetry. Also, let $C(p,0)$ be the focus, and let $D(a,0)$ be a point on $x$ axis such that $AD$ is perpendicular to $x$ axis. The vertex is $O(0,0)$, and let $E(p,e)$ where $e\gt 0$ be the intersection point of the parabola with the line perpendicular to $x$ axis passing through $C$.

(1)

Since the equation of the tangent line at $A$ is given by $by=2p(x+a)$, we have $B(-a,0)$, and so $$AC=\sqrt{(a-p)^2+(b-0)^2}=\sqrt{a^2-2ap+p^2+4pa}=\sqrt{(p+a)^2}=p+a=BC.$$

(2)

$OB=0-(-a)=a=OD$.

(3)

Solving $y^2=4px$ and $x=p$ gives $y=\pm 2p$, and so $e=2p$. Hence, $$\text{(the length of the latus rectum )}=2\times EC=2e=4p=4\times OC.$$

Let

• $A=$ focus of parabola $=(-1,-1)$
• $P=(7,13)$
• $G=$ foot of perpendicular from $A$ to tangent at $P$
• $V=$ vertex of parabola
• $a=AV$
• $\ell=AP=\sqrt{260}$
• $\theta=$ angle between $AP$ and $PG$.

The tangent at $P$ is $y=3x-8$ (given) which is the equation for $PG$.

Slope of $AP$, $m_1$ = $\dfrac74$.

Slope of $PG$, $m_2$ = $3$.

$$t=\tan\theta=\frac{m_2-m_1}{1+m_2m_1}=\frac{3-\frac74}{1+3\cdot \frac74}=\frac15$$

We make use of two properties of the parabola:

1. The tangent through $P$ bisects the angle between $AP$ and the line from $P$ perpendicular to the directrix.

2. $G$ lies on the tangent through $V$ (this line is parallel to the directrix).

to construct the diagram below

from which we can see that

\begin{align} AG=\frac a{\sin\theta}&=\ell\sin\theta\\ \ell&=\frac a{\sin^2\theta}=a \csc ^2\theta\\ &=a(1+\cot^2\theta)\\ &=a\left(1+\frac 1{t^2}\right)\\ &=26a &&\text{as } t=\frac 15\\ \ell^2=260&=26^2a^2\\ a^2&=\frac 5{13}\\ \text{Latus Rectum}=4a&=\color{red}{4\sqrt{\frac 5{13}}}\qquad\blacksquare \end{align}

Another method:

(This method does not require knowledge of the two properties above.
Instead the approach is to express $a$ in terms of $l$ and $\tan\theta$.)

Use the diagram and nomenclature as in the first method.
Consider the case for the parabola $y={x^2}/{(4a)}$.
Let $P=(2\lambda a, \lambda^2 a)$ be a point on this parabola.
Focus is $A (0,a)$.

$$\frac{dy}{dx}=\frac x{2a}=\lambda\qquad\text{at }P$$ Slope of $AP$, $m_2=\dfrac{\lambda^2a-a}{2\lambda a-0}=\dfrac\lambda 2-\dfrac 1{2\lambda}$
Slope of $PG$, $m_1=\lambda$. \begin{align} t=\tan\theta &=\frac{m_1-m_2}{1+m_1 m_2} =\frac{\lambda-\left(\frac \lambda2-\frac 1{2\lambda}\right)} {1+\lambda\left(\frac \lambda2-\frac 1{2\lambda}\right)} =\frac 1\lambda\\ \ell^2&=(\lambda^2a-a)^2+(2\lambda a-0)^2=a^2(\lambda^2+1)^2\\ \ell^2&=a^2\left(\frac 1{t^2}+1\right)\\ \end{align} This relationship applies for all parabolas.

For the given parabola, substitute values of $\ell^2$ and $t^2$: \begin{align} 260&=a^2(5^2+1)=26^2a^2\qquad\qquad\qquad\qquad\qquad\quad\\ a&=\sqrt{\frac 5{13}}\\ \text{Latus Rectum}=4a&=\color{red}{4\sqrt{\frac 5{13}}}\qquad\blacksquare \end{align}

• Will the solution be valid for any form of a parabola assumed apart from $y={x^2}/{(4a)}$?. Say I assumed $y^2=4ax$ and continued. – Ian Mar 19 at 3:10
• @Ian The length of the latus rectum of a parabola is 4 times its focal length. See also lexique.netmath.ca/en/latus-rectum and en.wikipedia.org/wiki/Conic_section. – Hypergeometricx Mar 20 at 7:09
• I have worked for $y^2=4ax$ and it yields same results. I appreciate your feedback. It is logical regardless of orientation the latus rectum is always the same – Ian Mar 20 at 13:00

Hint...rather than use the general form of the parabola you can use the geometric properties. This is an outline of an approach you could take, and you might like to work out the details for yourself.

Let the focus be $S(-1,-1)$ and $P(7,13)$. Let the directrix have equation $y=mx+c$, and let the foot of the perpendicular from $P$ to the directix be $N$.

Then the following properties of the parabola apply:

1. $PS=PN$ $$\Rightarrow \sqrt{260}=\left|\frac{13-7m-c}{\sqrt{1+m^2}}\right|$$

2. The latus rectum $L$ has length $4a$, where $2a$ is the perpendicular distance from the focus to the directrix. Therefore $$L=2\left|\frac{-1+m-c}{\sqrt{1+m^2}}\right|$$

3. The well-known reflector property of the parabola means that the tangent bisects the angle between $PS$ and $PN$. The gradient of $PS$ is $\frac 74$.

The angle between the tangent and $PS$ is $\theta$, where $$\tan\theta=\left|\frac{3-\frac 74}{1+3\times\frac 74}\right|=\frac15$$

Now let the gradient of $PN$ be $m_1$. You can use the same "angle between the gradients" formula to establish that $m_1=8\Rightarrow m=-\frac 18$.

It is then a matter of working out $c$ and then you get $L$.

I hope this helps.

Denote: F: Focus (-1,-1) P: The point on the parabola (7,13) M: the foot of the perpendicular from the Focus F on the tangent at P V: Vertex of Parabola

Then a standard property of the parabola (Prop 125 in Askwith - A Course of Pure Geometry) states that $FM^2 = FP.FV$. We easily obtain $FM = \sqrt {10}$ and $FP = 2 \sqrt{65}$

Then $FV = \sqrt{\frac{5}{13}}$.

The length of latus rectum $=4 FV = 4 \sqrt{\frac{5}{13}}$

I use some of the same properties, particularly the "optical" property, as the other responders here, but in a somewhat different way, so some of the same calculations will appear, but in a different guise. (This seems closest to hypergeometric's approach.)

There is a "similarity" property we can apply. For the "upward-opening" parabola with vertex at the origin, $\ y \ = \ \frac{1}{4p} \ x^2 \$ ($\ p \$ being the "focal distance", the distance from the vertex to the focus or to the directrix), so points on the curve can be characterised as $\ ( \ \pm \ 2 \ \sqrt{k} \ p \ , \ kp \ ) \$ . The slope of the tangent line to a point of the curve is $\ y' \ = \ \frac{1}{2p} \ x \$ , hence the slope at this point is $\ y' \ = \ \sqrt{k} \$ . We cannot use this immediately, however, as we can establish fairly quickly that the symmetry/focal axis of the parabola is not parallel to either coordinate axis. So we are faced with a bit of additional work.

We will make use of the "optical" property that the angle which a line parallel to the symmetry/focal axis makes to the tangent line to a point on the parabola is congruent to the angle that a line from the focus to that point makes to the same tangent line. Since we know nothing about the orientation of the parabola, we might use a scalar product of vectors to determine something about this angle. The vector from the focus $\ ( -1, \ -1 ) \$ to the given tangent point $\ (7, \ 13) \$ is $\ \langle \ 8, \ 14 \ \rangle \$ and the slope of the tangent line is $\ 3 \$ , which we may represent by a vector $\ \langle \ 1, \ 3 \ \rangle \$ . So we can compute

$$\cos \theta \ \ = \ \ \frac{\langle \ 8, \ 14 \ \rangle \ \cdot \ \langle \ 1, \ 3 \ \rangle }{ ( \sqrt{8^2 \ + \ 14^2} ) \ (\sqrt{1^2 \ + \ 3^2} ) } \ \ = \ \ \frac{50}{ \sqrt{260} \ \cdot \ \sqrt{10} } \ \ = \ \ \frac{5}{\sqrt{26}} \ .$$

We will similarly represent the slope of the symmetry axis by a vector $\ \langle \ 1, \ M \ \rangle \$ . We want the same acute angle between this vector and that for the tangent line, giving us

$$\cos \theta \ \ = \ \ \frac{\langle \ 1, \ 3 \ \rangle \ \cdot \ \langle \ 1, \ M \ \rangle }{ ( \sqrt{10} ) \ (\sqrt{1^2 \ + \ M^2} ) } \ \ = \ \ \frac{1 \ + \ 3M}{ ( \sqrt{10} ) \ (\sqrt{1^2 \ + \ M^2} ) } \ \ = \frac{5}{\sqrt{26}} \ .$$

This can be re-arranged into the quadratic equation $$\ 16 \ M^2 \ - \ 156 \ M \ + \ 224 \ = \ 4 \ M^2 \ - \ 39 \ M \ + \ 56 \ = \ 0 \ \ ,$$

with the solutions $\ M \ = \ \frac{39 \ \pm \ \sqrt{625}}{8} \ \ = \ \ 8 \ \ , \ \ \frac{7}{4} \$ . We want the steeper of these slopes, giving us the slope of the symmetry axis as $\ M \ = \ 8 \$ . (From this, we can develop some of mathlove's results, though we won't have need of those here.)

We see from this that the parabola is rotated "off the vertical" (by slightly over 7º clockwise, as it turns out), so we need to find the slope of the given tangent line relative to the symmetry axis in order to apply the aforementioned similarity property. The angle which the tangent line makes to the "horizontal" axis is given by $\ \tan \phi \ = \ 3 \$ (from its stated slope), so in a coordinate system for which the symmetry axis is "vertical" , we find the "transformed" slope from the "angle-addition formula" for tangent as

$$\ \tan \phi \ ' \ \ = \ \ \frac{3 \ + \ \frac{1}{8}}{1 \ - \ 3 \ \cdot \ \frac{1}{8}} \ \ = \ \frac{25}{5} \ \ = \ \ 5 \ \ ,$$

the $\ \frac{1}{8} \$ coming from the tangent of the clockwise angle that we must rotate the coordinate axes, which is the cotangent of the angle that the symmetry axis makes to the horizontal axis. The similarity property then tells us that $\ y' \ = \ \sqrt{k} \ = \ 5 \ \Rightarrow \ k \ = \ 25 \$ .

We still require the perpendicular distance of the tangent point from the symmetry axis. The line normal to that axis through $\ (7, \ 13) \$ is $\ y \ - \ 13 \ = \ -\frac{1}{8} \ ( x \ - \ 7 ) \ \Rightarrow \ y \ = \ \frac{111}{8} \ - \ \frac{1}{8} \ x \$ . The equation of the line containing the symmetry/focal axis is $\ y \ + \ 1 \ = \ 8 \ ( x \ + \ 1 )$ $\Rightarrow \ \ y \ = \ 8 \ x \ + \ 7 \$ , and these two lines intersect at $\ \left( \frac{11}{13} \ , \ \frac{179}{13} \right) \$ (as mathlove also finds).

At last, we apply the similarity property: the perpendicular distance from the symmetry axis to the tangent point is $\ 2 \ \sqrt{k} \ p \ \ = \ 10 \ p \$ . From the coordinates of the points, we obtain

$$(10 \ p)^2 \ \ = \ \ \left( \frac{11}{13} \ - \ 7 \right)^2 \ + \ \left( \frac{179}{13} \ - \ 13 \right)^2 \ \ = \ \ \left( \frac{11 \ - \ 91}{13} \right)^2 \ + \ \left( \frac{179 \ - \ 169}{13} \right)^2$$

$$\Rightarrow \ \ 100 \ p^2 \ \ = \ \ \left( \frac{-80}{13} \right)^2 \ + \ \left( \frac{10}{13} \right)^2 \ \ = \ \ \frac{6500}{13^2} \ \ \Rightarrow \ \ p \ = \ \frac{\sqrt{65}}{13} \ \ \text{or} \ \ \sqrt{\frac{5}{13}} \ \ .$$

With this information in hand, we could go on to find the location of the vertex of the parabola and the equation of its directrix, but none of that was requested. The length of the latus rectum of the parabola is $$\ 4 \ p \ = \ \frac{4 \ \sqrt{65}}{13} \ \ .$$

I am curious as to how much time and what resources were available in this "weekly test". While none of the calculations shown by the responders are terribly lengthy, providing a description of the techniques and formulas to be applied to the satisfaction of a "grader" -- particularly for a rotated conic section -- would seem to require a fair amount of writing.