# Focus of parabola with two tangents

A parabola touches x-axis at $(1,0)$ and $y=x$ at $(1,1)$. Find its focus.

My attempt : All I can say is that as angle subtended by this chord at focus is $90^\circ$ as angle between tangents is $45^\circ$. I can find equation of directrix by taking mirror image of focus in tangents and then use the fact that distance from focus is same as from directrix. But, this will get dirty.

Any help will be appreciated.

• do you mean y=x at (1, 1)? – Paul Apr 24 '14 at 17:35
• @Paul yes.${}{}$ edited – evil999man Apr 24 '14 at 17:37

(Sketch.) Let $P$ and $Q$ be two points on a parabola, and let $R$ be the point where the respective tangents to the parabola intersect. Let $X$ the midpoint of $PQ$. Then $RX$ is parallel to the axis of symmetry of the parabola (proved below). Draw lines through $P$ and $Q$ parallel to $RX$, and reflect these lines in the respective tangents; the focus $F$ is the intersection of the reflected lines.

To show that $RX$ is parallel to the axis of symmetry: Drop perpendiculars from $P,Q,R$ to the directrix, meeting it at $P',Q',R'$ respectively. As you alluded to, the tangent at $P$ is the perpendicular bisector of the segment $FP'$, and likewise for $Q$ and $FQ'$. So, in $\triangle FP'Q'$, two of the perpendicular bisectors pass through $R$; therefore the third does as well. Since $RR'$ is a line through $R$ and perpendicular to $P'Q'$, it must be the perpendicular bisector, that is, $R'$ is the midpoint of $P'Q'$. By parallels, (the extension of) $RR'$ bisects $PQ$, that is, $RR'$ passes through $X$. So $RX$ is perpendicular to the directrix, as claimed.

Edit: Just for reference, here's what this looks like analytically: The direction of $RX$ is $(2,1)$; reflecting in $RP$ just means exchanging $x$ and $y$ coordinates, so the direction of $PF$ is $(1,2)$, and the line through $P$ in that direction is $2x-y=1$. Reflecting in $RQ$ means negating the $y$-coordinate, so the direction of $QF$ is $(2,-1)$, and the line through $Q$ in that direction is $x+2y=1$. The intersection of these lines is $(\frac35,\frac15)$.

• I have to run right now, but let me know if more details are needed, and I'll expand when I return. – user21467 Apr 24 '14 at 18:43
• I mean, I'll expand my answer. I hope to remain the same size. – user21467 Apr 24 '14 at 18:44
• I don't see how RX is parallel to axis... maybe I will have to go through properties of conics again. Could you please provide a short geometry proof? – evil999man Apr 24 '14 at 19:16
• Added. I also changed $X$ to be the midpoint of $PQ$ (instead of the midpoint of the midline of $\triangle PQR$); this is simpler. – user21467 Apr 24 '14 at 21:30
• (+1) I rather like the idea of a "physical" approach to the parabola by using the "reflection of rays" at the tangent points in order to locate the focus. In practice, though, I wasn't convinced that the actual calculations would be much simpler than the straight-out use of analytic geometry and differentiation. – colormegone Apr 24 '14 at 22:47

In order to only be tangent at the specified points, the parabola cannot have its symmetry axis parallel to either coordinate axis. So we must make use of the general equation for a conic section, $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ . From the coordinates of the points that the parabola meets, we can conclude that

$$(1,0) : \ \ A + D + F \ = \ 0 \ \ \ , \ \ \ (1,1) : A + B + C + D + E + F \ = \ 0 \ \ \Rightarrow \ \ B + C + E \ = \ 0 \ \ .$$

By implicitly differentiating the conic section equation, we obtain

$$2Ax \ + \ B \ ( y + xy' ) \ + \ 2Cyy' \ + \ D \ + \ Ey' \ = \ 0 \ \ \Rightarrow \ \ y' \ = \ -\frac{2Ax + By+D}{Bx + 2Cy + E} \ \ .$$

At $\ (1,0) \$ , the slope of the parabola must match the slope of the $\ x-$ axis, thus,

$$y' \ = \ 0 \ = \ -\frac{2A+D}{B +E} \ \ \Rightarrow \ \ 2A + D \ = \ 0 \ \ \Rightarrow \ \ A - F \ = \ 0 \ \ ;$$

at the other point of tangency, $\ (1,1) \$ , the slope of the parabola is the same as the slope of the line $\ y = x \$ , so

$$y' \ = \ 1 \ = \ -\frac{2A+B+D}{B+2C +E} \ \ \Rightarrow \ \ 2A + B + D \ = \ -(B+2C+E)$$

$$\Rightarrow \ \ 2A + 2B + 2C + D + E \ = \ 0 \ \ .$$

Applying earlier results, we find that $\ 2B + 2C + E \ = 0 \ \ \Rightarrow \ \ B + C \ = \ 0 \ \ , \ \ E = 0 \$ . We have, to this point, reduced the conic section equation to $Ax^2 + Bxy - By^2 - 2Ax + A = 0 \$ .

We still need a fifth equation, which we can obtain from the discriminant for the general conic section: for a parabola, $\ B^2 - 4AC \ = \ 0 \$ . For our parabola, then, $\ B^2 \ - \ 4A \ (-B) \ = \ 0$ $\Rightarrow \ \ B \ (B + 4A) \ = \ 0 \$ . We cannot have $\ B = 0 \$ for a rotated conic section, so $\ B \ = \ -4A \$ , making the equation $$Ax^2 - 4Axy + 4Ay^2 - 2Ax + A = 0 \ \ \Rightarrow \ \ A \ (x^2 - 4xy + 4y^2 - 2x + 1) \ = \ 0 \ \ .$$

A can then be set equal to one.

The angle of rotation of the parabola is given by

$$\ \tan(2 \theta) \ = \ \frac{B}{A-C} \ = \ \frac{-4A}{A - 4A} \ = \ \frac{4}{3} \ = \ \frac{2 \ \tan \theta}{1 \ - \ \tan^2 \theta}$$

$$\Rightarrow \ \ \tan \theta \ = \ \frac{1}{2} \ , \ (-2) \ \ .$$

This gives the slope of the symmetry axis for the parabola; since the parabola "opens" into the first quadrant, we adopt the positive slope solution, so the vertex and focus line along a line of slope $\ \frac{1}{2} \ .$

The above is a graph of what we have to this point. It remains to determine the location of the focus along the symmetry axis. For this, we will use the properties of the latus rectum of the parabola, which passes through the focus $\ F \$ and is perpendicular to the symmetry axis.

We will label the coordinates of the focus $\ F \ (X,Y) \$ , those of the endpoints of the latus rectum $\ A \ (x_A , y_A) \ \ \text{and} \ \ B \ (x_B , y_B) \$ , and those of the vertex $\ V \ (x_V, y_V ) \$ . The focal distance $\ p \$ is the distance from $\ F \$ to $\ V \$ ; the distance from $\ F \$ to either $\ A \$ or $\ B \$ is $\ 2p \$ .

As the slope of the symmetry axis is $\ \frac{1}{2} \$ , the slope of the latus rectum is then $\ -2 \$ . By application of the "distance formula" or trigonometry, we determine that

$$X \ - \ x_V \ = \ \frac{2}{\sqrt{5}}p \ \ , \ \ Y \ - \ y_V \ = \ \frac{1}{2} \ (X \ - \ x_V ) \ = \ \frac{1}{\sqrt{5}}p \ \ ;$$

$$X \ - \ x_A \ = \ \frac{2}{\sqrt{5}}p \ \ , \ \ Y \ - \ y_A \ = \ -2 \ (X \ - \ x_A ) \ = \ -\frac{4}{\sqrt{5}}p \ \ ; \ \ \text{and}$$

$$X \ - \ x_B \ = \ -\frac{2}{\sqrt{5}}p \ \ , \ \ Y \ - \ y_B \ = \ -2 \ (X \ - \ x_B ) \ = \ \frac{4}{\sqrt{5}}p \ \ .$$

We can now use our result for the conic section equation, $\ x^2 - 4xy + 4y^2 - 2x + 1 \ = \ 0 \$ , to write

$$x_V^2 \ - \ 4 \ x_V \ y_V \ + \ 4y_V^2 \ - \ 2x_V \ + \ 1$$ $$= \ \left( X - \frac{2}{\sqrt{5}}p \right)^2 \ - \ 4 \ \left( X - \frac{2}{\sqrt{5}}p \right) \ \left( Y - \frac{1}{\sqrt{5}}p \right) \ + \ 4\left( Y - \frac{1}{\sqrt{5}}p \right)^2 \ - \ 2\left( X - \frac{2}{\sqrt{5}}p \right) \ + \ 1 \ = \ 0 \ \ ;$$

$$x_A^2 \ - \ 4 \ x_A \ y_A \ + \ 4y_A^2 \ - \ 2x_A \ + \ 1$$ $$= \ \left( X - \frac{2}{\sqrt{5}}p \right)^2 \ - \ 4 \ \left( X - \frac{2}{\sqrt{5}}p \right) \ \left( Y + \frac{4}{\sqrt{5}}p \right) \ + \ 4\left( Y + \frac{4}{\sqrt{5}}p \right)^2 \ - \ 2\left( X - \frac{2}{\sqrt{5}}p \right) \ + \ 1 \ = \ 0 \ \ ; \ \ \text{and}$$

$$x_B^2 \ - \ 4 \ x_B \ y_B \ + \ 4y_B^2 \ - \ 2x_B \ + \ 1$$ $$= \ \left( X + \frac{2}{\sqrt{5}}p \right)^2 \ - \ 4 \ \left( X + \frac{2}{\sqrt{5}}p \right) \ \left( Y - \frac{4}{\sqrt{5}}p \right) \ + \ 4\left( Y - \frac{4}{\sqrt{5}}p \right)^2 \ - \ 2\left( X + \frac{2}{\sqrt{5}}p \right) \ + \ 1 \ = \ 0 \ \ .$$

EDIT -- We should now have sufficient information to solve for the coordinates of the focus $\ (X,Y) \$ and the focal distance $\ p \$ . Setting successive pairs of these last three equations equal produces

$$5p \ - \ \frac{5}{\sqrt{5}} X \ + \ \frac{10}{\sqrt{5}}Y \ = \ 0 \ \ , \ \ \ \mathbf{[1]}$$

$$5p \ + \ \frac{5}{\sqrt{5}} X \ - \ \frac{10}{\sqrt{5}}Y \ = \ \frac{2}{\sqrt{5}} \ \ , \ \ \ \mathbf{[2]}$$

$$\ \frac{5}{\sqrt{5}} X \ - \ \frac{10}{\sqrt{5}}Y \ = \ \frac{1}{\sqrt{5}} \ \ . \ \ \ \mathbf{[3]}$$

Adding equation (1) to equation (2) yields $\ 10p \ = \ \frac{2}{\sqrt{5}} \ \ \Rightarrow \ \ p \ = \ \frac{\sqrt{5}}{25} \ \approx \ 0.0894 \$ . These three equations turn out to be linearly dependent [subtracting (1) from (2) gives twice (3) ] . Dividing (3) through by $\ -2\sqrt{5} \$ leads to $\ Y \ = \ \frac{1}{2}X \ - \frac{1}{10} \$ , which tells us that the equation of the symmetry axis is $\ y \ = \ \frac{1}{2}x \ - \frac{1}{10} \$ .

Since the vertex is also on the symmetry axis, we can apply

$$x_V^2 \ - \ 4 \ x_V \ y_V \ + \ 4y_V^2 \ - \ 2x_V \ + \ 1$$ $$= \ x_V^2 \ - \ 4 \ x_V \ \left(\frac{1}{2}x_V \ - \frac{1}{10}\right) \ + \ 4 \ \left(\frac{1}{2}x_V \ - \frac{1}{10}\right)^2 \ - \ 2x_V \ + \ 1$$ $$= \ \frac{4}{100} \ - \ 2 x_V \ + \ 1 \ = \ 0 \ \ \Rightarrow \ \ x_V \ = \ \frac{13}{25} \ \ \Rightarrow \ \ y_V \ = \frac{4}{25} \ \ .$$

Finally, the coordinates of the focus are found from

$$X \ = \ x_V \ + \frac{2}{\sqrt{5}} \cdot \frac{\sqrt{5}}{25} \ = \ \frac{13 \ + \ 2}{25} \ = \ \frac{3}{5} \ \ ,$$ $$Y \ = \ y_V \ + \ \frac{1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{25} \ = \ \frac{4 \ + \ 1}{25} \ = \ \frac{1}{5} \ \ .$$

We see from this that it would not have been a simpler approach to work out the details in a coordinate system aligned with one of the coordinate axes, and then to rotate the parabola appropriately, as the tangent points are not placed symmetrically to either side of the parabola's symmetry axis.

I have a different approach to this problem. Using the general conic equation

$$A x^2 + B x y + C y^2 + D x + E y + F =0$$

and its derivative (for tangents)

$${\rm d} x ( 2 A x + B y + D ) + {\rm d} y ( B x + 2 C y + E) =0$$

With the parabola constraint $B^2 = 4 A C$, and the following 4 constraints, all coefficients can be found

\begin{align} A + D + F & =0 & & \mbox{curve} & (x=1,y=0) \\ A + B + C + D +E + F & = 0 & & \mbox{curve} & (x=1,y=1) \\ 2 A + D & =0 & & \mbox{tangent} & (x=1,y=0,{\rm d}x=1,{\rm d}y=0) \\ 2 A + 2 B+ 2 C + D + E & =0 & & \mbox{tangent} & (x=1,y=1,{\rm d}x=1,{\rm d}y=1) \\ \end{align}

The above is solved for $$C=-B\\D=-2 A\\E=0\\F=A$$

$$A x^2 -2 A x - B y^2 + B x y +A =0$$ and $B^2 = 4 A C = -4 A B \} B = -4 A$

$$A \left( x^2 - 4 x y - 4 y^2 - 2 x + 1 \right) = 0 \\ A = 1 \mbox{ arbitrary}$$

To find the center, the quadratic is at an extrema ( slope is zero ) at

$$2 A x_c + B y_c +D =0 \\ B x_c + 2 C y_c +E =0$$

$$x_c = \frac{2 C D - B E}{B^2 - 4 A C} = \frac{4 A}{4 A +B} \\ y_c = \frac{2 A E - B D}{B^2 - 4 A C} = \frac{2 A}{4 A +B}$$

with solution for the parabolic constraint $B=0$ yielding

$$x_c = 1 \\ y_c = \frac{1}{2}$$

• Unlike the other answers, my system of equations are all linear, and I do not use any trigonometry. – ja72 Apr 25 '14 at 0:55
• Could you please clarify what you are doing in your last paragraph? What is $\ (x_c , y_c) \$ , since it is not the focus of the parabola? – colormegone Apr 25 '14 at 6:05
• My bad. This is the center of the conic (See point X in the accepted answer). – ja72 Apr 25 '14 at 17:24