<PRE><B>Question 1187647</B>
&lt;br&gt;
The given information is unusual, so initially we don&#39;t have a good idea of where to go with the problem.  So we do the only thing we can -- write the general form of the equation and use each the two given points in that equation to see what it gives us.&lt;br&gt;
The vertex is on the line y=2x, so we can call the coordinates of the vertex (a,2a).  Then the vertex form of the equation of a parabola parallel to the x-axis with vertex (a,2a) is&lt;br&gt;
{{{(x-a) = (1/(4p))(y-2a)^2}}}&lt;br&gt;
Plug in (x,y)=(1.5,1) and (x,y)=(3,4) to get two equations in a and p:&lt;br&gt;
{{{1.5-a=(1/(4p))(1-2a)^2}}}&lt;br&gt;
{{{4p(1.5-a)=1-4a+4a^2}}} [1]&lt;br&gt;
{{{3-a=(1/(4p))(4-2a)^2}}}&lt;br&gt;
{{{4p(3-a)=16-16a+4a^2}}} [2]&lt;br&gt;
Subtract [1] from [2]:&lt;br&gt;
{{{4p(1.5)=15-12a}}}
{{{6p=15-12a}}}
{{{p=2.5-2a}}}
{{{4p=10-8a}}} [3]&lt;br&gt;
Substitute [3] in [2]:&lt;br&gt;
{{{(10-8a)(3-a)=16-16a+4a^2}}}
{{{30-34a+8a^2=16-16a+4a^2}}}
{{{4a^2-18a+14=0}}}
{{{2a^2-9a+7=0}}}
{{{(2a-7)(a-1)=0}}}&lt;br&gt;
{{{a=3.5}}} or {{{a=1}}}&lt;br&gt;
There will be two parabolas that satisfy the given conditions.&lt;br&gt;
(1) a=3.5; 4p=10-8a=-18; 2a=7&lt;br&gt;
{{{x-3.5=(-1/18)(y-7)^2}}}&lt;br&gt;
(2) a=1; 4p=10-8a=2; 2a=2&lt;br&gt;
{{{x-1=(1/2)(y-2)^2}}}&lt;br&gt;
Here is a graph of the two parabolas, one with vertex (1,2) and the other with vertex (3.5,7) and both passing through the points (1.5,1) and (3,4)&lt;br&gt;
{{{graph(400,400,-2,6,-2,10,sqrt(2(x-1))+2,-sqrt(2(x-1))+2,sqrt(-18(x-3.5))+7,-sqrt(-18(x-3.5))+7)}}}</PRE>