Parabola

Feb 2010
4
1
1.Prove that the orthocentre of the triangle formed by the three tangents to a parabola lie on a directrix.

2.If the Latus Rectum= \(\displaystyle 4\), the vertex is \(\displaystyle (-2,0)\) and the equation of the axis is \(\displaystyle 3x+4y+6=0\) then find the equation of a parabola satisfying these. (Ans: \(\displaystyle 9x^2+24xy+16y^2-44x+108y-124=0,\
9x^2+24xy+16y^2+36x+48y+108=0\))

3.Show that circle described on focal chord of a parabola as diameter touches its directrix.

Kindly please help in the above questions. Any thoughts or hints would be highly appreciated. Thank you. (Happy)

Question 3 has been solved. Thanks to Soroban! (Happy)
 
Last edited:

Soroban

MHF Hall of Honor
May 2006
12,028
6,341
Lexington, MA (USA)
Hello, riemann!

3.Show that circle described on focal chord of a parabola as diameter touches its directrix.

Let the parabola be: .\(\displaystyle x^2 \:=\:4py\)

This parabola opens upward; its vertex is at the Origin.
The focus is at \(\displaystyle (0,p)\)
The directrix is: .\(\displaystyle y \:=\:p\)
The focal chord (latus rectum) has length \(\displaystyle 4p.\)

The graph looks like this:


Code:
                |
      *         |         *
                |
                |
       *        |        *
               F|(0,p)
(-2p,p) o - - - o - - - o (2p,p)
         *      |      *
           *    |    *
    ------------*--------------
                |
                |
        - - - - + - - - - -   y = -p
                |
                |

The circle has center \(\displaystyle (0,p)\) and radius \(\displaystyle 2p\).

You should be able to complete the problem now.

 
  • Like
Reactions: riemann

Opalg

MHF Hall of Honor
Aug 2007
4,039
2,789
Leeds, UK
1.Prove that the orthocentre of the triangle formed by the three tangents to a parabola lie on a directrix.
For the parabola \(\displaystyle y^2=4ax\), the tangent at the point \(\displaystyle (ap^2,2ap)\) has equation \(\displaystyle x-py+ap^2 = 0\). Check that the tangents at the points \(\displaystyle (ap^2,2ap)\) and \(\displaystyle (aq^2,2aq)\) meet at the point \(\displaystyle (apq, a(p+q))\). The line through this point perpendicular to the tangent at the point \(\displaystyle (ar^2,2ar)\) will have gradient \(\displaystyle -r\), and its equation is therefore \(\displaystyle y-a(p+q) = -r(x-apq)\).

That line is one of the altitudes of the triangle formed by the tangents at the three points on the parabola. It meets the directrix \(\displaystyle x=-a\) at the point given by \(\displaystyle y-a(p+q) = -r(-a-apq)\), or \(\displaystyle y = a(p+q+r+pqr)\). That last equation is symmetric in p, q and r. So both the other altitudes of the triangle will pass through the same point on the directrix. In other words, the orthocentre of the triangle lies on the directrix.