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This is a question I found:

A tangent is drawn to a parabola at the point $(12,6)$ to meet the directrix at $\left( 2,{8\over 3} \right)$. If focus lies on the x axis then what is the equation of the parabola?

From this it can be deduced that directrix is $x=2$, and that the whole parabola along with directrix are located in the first quadrant. Therefore the origin is certainly not at the vertex. The coordinates of the focus are given in the solution as $(4,0)$, but how can I get them? If the vertex had been the origin I could have used the fact that the distance from the vertex to the directrix is equal to the focal length, but here I don't know what to do. How do I approach this problem?

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  • $\begingroup$ The interesting bit here (relative to my personal limited knowledge) is, how we know that this is a parabola symmetric to the $x$-axis, and not some more funny parabola (rotated, shifted etc), starting from the general conic section $ax^2+bxy+cy^2 +dx + ey + f = 0$ and the construction rule parabola = all points with same distance to focus point and directrix line plus the bits of information in the task. $\endgroup$ – mvw May 20 '18 at 11:12
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The equation of parabola is $$y^2=2p(x-a)$$ so the focus is at $F(a+{p\over 2},0)$ and the equation of tangent is $x-3y+6=0$.

Since a general equation of tangent at $T(x_0,y_0)$ if $y_0\ne 0$ is $$y={p\over y_0}x +p{x_0-2a\over y_0}$$ we get ${p\over y_0} ={1\over 3}$ and $p(x_0-2a)=2y_0$, so $p=2$ and $a= 3$, so parabola is $$y^2=4(x-3)$$


Or you can do like this:

Since ${1\over 3} = {p\over 6}$ we get $p=2$ so $F(a+1,0)$. Let $T'(2,6)$ be orthogonal projection on directrix. Now the midpoint $M({a+3\over 2},3)$ of a segment $T'F$ is on tangent, so ${a+3\over 2}-9+6=0$ so $a=3$.

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    $\begingroup$ I know this is a silly doubt but won't the general equation for tangent be $ yy_0 = p(x+x_0) - 2pa $ or $ y = (p/y_0)x + (p/y_0)(x_0-2a) $? $\endgroup$ – Hema May 20 '18 at 10:44
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    $\begingroup$ Yes, I made a correction $\endgroup$ – Aqua May 20 '18 at 10:48
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You could use the reflective property of parabolas: A normal to the parabola at $(12,6)$ is $(-(6-8/3),12-2)=(-10/3,10)$ and so the reflection of the direction vector $(1,0)$ of the parabola’s axis from the tangent line is (eliminating the common factor of $10$) $$(1,0)-{(1,0)\cdot(-1,3)\over(-1,3)\cdot(-1,3)}(-1,3) = \left(\frac45,\frac35\right).$$ The intersection of the line through $(12,6)$ that has this direction vector with the $x$-axis is the parabola’s focus, $(4,0)$.

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