# Find the equation of parabola, when tangent at two points and vertex is given

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.

• 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.)
– Blue
May 14, 2018 at 6:39
• 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. May 15, 2018 at 11:33

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 $D$.

For this I will use the property that in any parabola, foot of perpendicular from focus upon any tangent lies on the tangent at vertex. This is a general result and I will prove it for the standard $${y^2}=4ax$$ parabola. The focus is S($$a,0$$), tangent at vertex is $$x=0$$. Now assume a point P($$a{t^2},2at$$). The tangent at point P is $$ty=x+a{t^2}$$. This intersects $$x=0$$ at M($$0,at$$). Now slope of SM =$$\left(\frac{at-0}{0-a}\right)$$=$$-t$$.

Also slope of tangent PM is $$1/t$$. Multiply both slopes and you get -1 hence they are perpendicular.

Using the same property, I find the intersection of both tangents on the tangent at vertex. I assumed focus as S($$h,k$$)

$$3x+4y-7=0$$ and $$x-y=0$$ gives solution M$$(1,1)$$. Slope of SM x Slope of PM=$$-1$$

$$\left(\frac{1-k}{1-h}\right)* \left(\frac{-3}{4}\right)$$=-1. This gives $$3-3k=4-4h$$ or $$1+3k-4h=0$$.

Similarly the tangent at Q intersects $$x-y=0$$ at N(2,2). Slope SN x Slope QN=$$-1$$

$$\left(\frac{2-k}{2-h}\right)$$ * $$\left(\frac{-2}{3}\right)$$=$$-1$$. This gives $$4-2k=6-3h$$ or $$2+2k-3h=0$$.

Solving $$1+3k-4h=0$$ and $$2+2k-3h=0$$ gives $$k=5$$,$$h=4$$. Thus your focus is $$S(4,5)$$. Distance of focus from tangent at vertex using perpendicular distance formula $$\left(\frac{|4-5|}{\sqrt{1^2+1^2}}\right)$$ =1/$$\sqrt{2}$$ Which is also the value for '$$a$$'. Latus rectum= $$4a$$ = $$4 * 1/\sqrt{2}$$ = $$2\sqrt{2}$$.

Also axis is perpendicular to tangent at vertex $$x-y=0$$. Its slope would be $$-1$$ and it passes through S($$4,5$$). Hence the required equation is $$x+y-9=0$$.

Now solving $$x+y-9=0$$ and $$x-y=0$$ we get vertex at A($$9/2,9/2$$).