Focus is $(1,1)$ and equation of the tangent is $x+y=1$. With this information, we have to find equation of parabola and length of latus rectum. (Solution is given as $x² -2xy+y² -4x -4y+4=0$)

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    $\begingroup$ Do we assume parabola "lined up" with coordinate axes? $\endgroup$ – coffeemath Feb 27 '17 at 16:58
  • $\begingroup$ No information for the same is given. $\endgroup$ – Shoaib Ashraf Feb 27 '17 at 17:05
  • $\begingroup$ With no more information, there are it seems many possible parabolas with that focus and having that as one of the tangent lines. Does the problem just say to find one such parabola, or find all such? $\endgroup$ – coffeemath Feb 27 '17 at 17:11
  • $\begingroup$ @coffeemath I've added the solution. $\endgroup$ – Shoaib Ashraf Feb 27 '17 at 17:16
  • $\begingroup$ @coffeemath From the question’s title, I’d say no. Note, too, that the question’s title says that the given tangent is at the vertex. $\endgroup$ – amd Feb 27 '17 at 20:22

A parabola’s vertex is midway between its focus and directrix. You’re given the tangent line at the vertex, which is parallel to the directrix, so it should be a straightforward matter to find an equation for the directrix: you have its slope/normal and you can find a point on it by reflecting the given focus point $(x_f,y_f)$ in the tangent line. If you put the resulting equation in the form $ax+by+c=0$, then an equation for the parabola is $$(x-x_f)^2+(y-y_f)^2={(ax+by+c)^2\over a^2+b^2}.$$ This equation just expresses the definition of a parabola as the set of points that are equidistant from the focus and directrix. Rearrange this into whatever form is required.

As for the latus rectum, you don’t even need the equation of the parabola to compute its length. You can get the focal distance from the information given, and the latus rectum length is a fixed multiple of this distance.

  • $\begingroup$ Is it in the problem that the given tangent line goes through the vertex? $\endgroup$ – coffeemath Feb 28 '17 at 0:21
  • $\begingroup$ @coffeemath It’s in the question’s title. Without that extra information, the problem doesn’t have a unique solution, of course. $\endgroup$ – amd Feb 28 '17 at 1:33

Got the solution on another thread, thanks to Shashank. Find slope of the given line (tangent). Say it is m. So the slope of line (axis) joining the point and its mirror image is -1/m. Use slope point form to find equation of the line (axis) and find its interaection with given line (tangent). Finally use the intersection point (vertex) in midpoint formula to get the required point. Original post

  • $\begingroup$ Seems like this might be more work than necessary. The equation of the directrix is going to be $x+y=c$ for some c. Choose $c$ so that the distance of the focus from this line is twice the distance of the focus from the tangent line. Also, what would you do if the tangent was vertical? You don’t have a value for $m$ in that case. $\endgroup$ – amd Feb 28 '17 at 4:57


If the equation of the directrix is $ax+by+c=0$

the equation of the required parabola: $$(x-1)^2+(y-1)^2=\dfrac{(ax+by+c)^2}{a^2+b^2}$$

Put $y=1-x$ to find a quadratic equation in $x,$ whose roots represent the abscissa of the intersection.

Now for tangency, both roots will be same.

Can you take it from here?

  • $\begingroup$ How can given tangent intersect with Directrix? It passes through vertex which makes it parallel to Directrix. $\endgroup$ – Shoaib Ashraf Feb 27 '17 at 17:44
  • $\begingroup$ @ShoaibAshraf, How do you calculate the vertex? $\endgroup$ – lab bhattacharjee Feb 27 '17 at 17:45
  • $\begingroup$ Intersection point of parabola and axis of parabola. $\endgroup$ – Shoaib Ashraf Feb 27 '17 at 17:47
  • $\begingroup$ @ShoaibAshraf, What is the axis of the parabola in question? $\endgroup$ – lab bhattacharjee Feb 27 '17 at 17:50
  • $\begingroup$ I cannot find one. $\endgroup$ – Shoaib Ashraf Feb 27 '17 at 17:51

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