Given parabola $y^2=4x$ find the equation of the chord with $(2,3)$ as its midpoint 
I have been given a parabola whose equation is $y^2=4x$. The question asks to find the equation of the chord with $(2,3)$ as its midpoint.

I tried to solve the problem using parametric coordinates for a parabola. Taking the two ends of the chord as $(t^2,2t)$ and $(s^2,2s)$, I got the following relations between $t$ and $s$:
$$t^2+s^2=4 \quad\text{and}\quad t+s=3 \tag{1}$$
Using $$(t+s)^2 +(t-s)^2=2(t^2+s^2) \tag{2}$$ 
I got 
$$(t-s)^2=-1 \tag{3}$$ 
After this I concluded that such a chord is not possible. But somewhere else I found that equation of such a chord is given by $3y-2x=5$. 
When I tried to find the points of intersection of this chord with the parabola, I got imaginary points. 

Is such a chord possible?

 A: The point $(2,3)$ isn’t in the interior of the parabola, so no chord at all passes through that point. You’re correct in that regard.  
On the other hand, the chord of a parabola that is bisected by a given interior point is parallel to the polar of that point. Applying this criterion the the point $(2,3)$, we can find find that its polar is $3y=2(x+2)$ and the parallel line to this through $(2,3)$ is $3y-2x=5$. This is exactly the line that you found somewhere else. Perhaps that source blindly computed this line using something like the above method without bothering to check that the point was in the interior in the first place.
A: Let the equation of the chord be $$m=\dfrac{y-3}{x-2}\iff y=mx+3-2m$$
Let us find the intersection $$4x=(mx+3-2m)^2\iff m^2x^2+2x(m(3-2m)-2)+(3-2m)^2=0$$
which is a quadratic equation in $x$
So, $$2=\dfrac{\dfrac{2(2m^2-3m+2)}{2\cdot m^2}}2$$
A: The problem has solution only if $(2,3)\in (y^2-4a x < 0)$
so assuming $a > \frac{9}{8}$ and making the intersection between
$y^2-4ax = 0$ and $y = 3 + m(x-2)$ we obtain two solutions
$$
y_{1,2} = \frac{\frac{9}{4}\pm\frac{3 \sqrt{m (2
   m-3)+\frac{9}{8}}}{\sqrt{2}}}{m}
$$
now equating
$$
y_2 - 3 = 3 - y_1
$$
we get $m = \frac{2a}{3}$
