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The equation of tangent at the point $P$, $Q$ and vertex $A$ of a parabola are $3x+4y-7=0, 2x+3y-10=0$, and $x-y=0$ respectively. I'm trying to show:

  1. Focus is $(4,5)$
  2. Length of Latus Rectum is $2\sqrt2$
  3. Axis is $x+y-9=0$
  4. Vertex is $(\frac{9}{2}$,$\frac{9}{2})$

I encountered this is a Multiple Choice question where all the answers are correct. My approach was the following:

at vertex $A$, the equation of tangent is $x-y=0$, hence equation of normal is $x+y=t$. This is the axis so focus should pass through it. Let focus be $(4,5)$ then the equation of axis is $x+y-9=0$

Upon solving simultaneous equation, we get the vertex as $(\frac{9}{2}$,$\frac{9}{2})$, and the distance from focus to vertex is $\frac{1}{\sqrt2}$ which when multiplied by 4 is equal to latus-rectum $2\sqrt2$.

Though I got the answers correct but my approach is wrong as I need to find the answer by calculation and not by plugging the value and see the which answers are consistent.

I seem to recall that the image of focus on any tangent line lies on the directix of the parabola. Thus if we take focus as $(a,b)$ we can find three points on the directix. The slope will be parallel to $x-y=0$, and what's perpendicular to directrix will be the axis of the parabola.

However I'm not able to find the correct answer this way.

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  • $\begingroup$ You have the right idea. Reflect $(a,b)$ in the three lines to get, say, $H$, $J$, $K$. Then the fact that $\overleftrightarrow{HJ}$ and $\overleftrightarrow{HK}$ are parallel to $x-y=0$ implies $(H-J)\cdot(1,-1) = 0$ and $(H-K)\cdot(1,-1) = 0$. That's a system of two equations that you can solve for $a$ and $b$. (I think it'd be fair at that stage to simply "observe" that $(a,b)=(4,5)$ satisfies the system.) $\endgroup$
    – Blue
    May 14, 2018 at 6:39
  • $\begingroup$ Geometrically speaking, the triangle formed by the three tangents has a twofold significance: Its orthocenter is contained in the parabola's directrix (by Steiner), and its circumcircle contains the focus (by Lambert). Indeed the focus is the intersection of the mirror images of the directrix across the tangents. You also know that the directrix is parallel to the vertex tangent. That is enough for construction and/or calculation. $\endgroup$
    – ccorn
    May 15, 2018 at 11:33

1 Answer 1

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To find the vertex of that parabola, you can make use of a nice general result:

given three tangents of a parabola, if $AB$ is that part of a tangent which is comprised between the other two tangents (see diagram), then the projection of $AB$ onto a line perpendicular to the axis has a fixed length, independent of the position of $AB$.

In our particular case, if $V$ (vertex) and $D$ are the tangency points of the "outer" tangents, $C$ is their intersection point, and $B'$, $D'$ are the projections of $B$, $D$ on tangent $VC$ (which is perpendicular to the axis), we obtain from the above result:

$$AB'=CD'=VC.$$

As points $A$, $B$, $C$, $B'$ can be readily found from the data, one can also easily find vertex $V$ and tangency point $D$.

enter image description here

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