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

Question

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.

Answer 1

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 $D$.

Answer 2

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($9/2,9/2$).