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Question: A parabola is drawn such that each vertex of a given triangle is the pole of the opposite side; show...

A parabola is drawn such that each vertex of a given triangle is the pole of the opposite side; show that the focus of the parabola lies on the nine point circle of the triangle and that the orthocentre of the triangle formed by joining the middle points of the sides line on the directrix.

Explanation

Solution

Hint: The circle circumscribing the triangle is joined by the middle point of the sides forming a 99 point circle of triangle. Use the mid-point theorem and form the equation of line. By simplifying you will get an equation similar to y=mx+cy=mx+c.

Complete step-by-step solution -
The circle circumscribing the triangle is formed by joining the middle points of the sides of six of the given triangle and this will form a 99 point circle of the triangle.
The circle circumscribing the triangle is formed by 33 tangent on the parabola, which always passes through the focus of parabola.

In geometry, the 99point circle is a circle that can be constructed for any given triangle.
The 99 points are
(i) Midpoint of each side of the triangle
(ii) The foot of each attitude
(iii) The mid point of the line segment from each vertex of the triangle to the orthocentre.
Now in this question we have to prove that the lines joining the midpoints of the given triangle are tangents on the parabola.
We know the equation of parabola ⇒y2=4ax.................(1)\Rightarrow {{y}^{2}}=4ax.................\left( 1 \right)
The given triangle is ABCABC. Let PQPQ be the chord of contact of AA w.r.t to equation (1)\left( 1 \right)
Let BCBC be the polar of AA w.r.t equation (1)\left( 1 \right)
∴BC\therefore BC and PQPQ must lie on the same line.
i.e. the line joining the mid points of APAP and AQAQ also passes through the mid point of ABAB and ACAC
Let us consider the coordinates of PP and QQ as (at12,2at1)\left( a{{t}_{1}}^{2},2a{{t}_{1}} \right) and (at22,2at2)\left( a{{t}_{2}}^{2},2a{{t}_{2}} \right) respectively, so that the coordinates of the point of intersection of the tangents PP and QQare [at1t2,a(t1+t2)]\left[ a{{t}_{1}}{{t}_{2}},a\left( {{t}_{1}}+{{t}_{2}} \right) \right]
∴\therefore Mid point of AP⇒where A(at12x1 ,2at1y1 )AP\Rightarrow \text{where }A\left( \underset{{{x}_{1}}}{\mathop{a{{t}_{1}}^{2}}}\,,\underset{{{y}_{1}}}{\mathop{2a{{t}_{1}}}}\, \right)
P(at1t2x2 ,a(t1+t2)y2 )P\left( \underset{{{x}_{2}}}{\mathop{a{{t}_{1}}{{t}_{2}}}}\,,\underset{{{y}_{2}}}{\mathop{a\left( {{t}_{1}}+{{t}_{2}} \right)}}\, \right)
Formula of mid point theorem (x1+x22,y1+y22)\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)
∴\therefore Midpoint of AP⇒[at12+at1t22,2at1+a(t1+t2)2]AP\Rightarrow \left[ \dfrac{a{{t}_{1}}^{2}+a{{t}_{1}}{{t}_{2}}}{2},\dfrac{2a{{t}_{1}}+a\left( {{t}_{1}}+{{t}_{2}} \right)}{2} \right]
Simplifying the above coordinate =\left\\{ \dfrac{a{{t}_{1}}\left( {{t}_{1}}+{{t}_{2}} \right)}{2},\dfrac{a\left( 3{{t}_{1}}+{{t}_{2}} \right)}{2} \right\\}
Similarly the coordinates of midpoint of AQAQ
A(at22x1 ,2at2y1 ) and Q(at1t2x2 ,2(t1t2)y2 )A\left( \underset{{{x}_{1}}}{\mathop{a{{t}_{2}}^{2}}}\,,\underset{{{y}_{1}}}{\mathop{2a{{t}_{2}}}}\, \right)\text{ and Q}\left( \underset{{{x}_{2}}}{\mathop{a{{t}_{1}}{{t}_{2}}}}\,,\underset{{{y}_{2}}}{\mathop{2\left( {{t}_{1}}{{t}_{2}} \right)}}\, \right)
Midpoint of AQ=\left\\{ \dfrac{a{{t}_{2}}\left( {{t}_{1}}+{{t}_{2}} \right)}{2},\dfrac{a\left( {{t}_{1}}+3{{t}_{2}} \right)}{2} \right\\}
(x1y1)=[at1(t1+t2)2,a(3t1+t2)2]..................(2) (x2y2)=[at2(t1+t2)2,a(t1+3t2)2]...................(3) \begin{aligned} & \left( {{x}_{1}}{{y}_{1}} \right)=\left[ \dfrac{a{{t}_{1}}\left( {{t}_{1}}+{{t}_{2}} \right)}{2},\dfrac{a\left( 3{{t}_{1}}+{{t}_{2}} \right)}{2} \right]..................\left( 2 \right) \\\ & \left( {{x}_{2}}{{y}_{2}} \right)=\left[ \dfrac{a{{t}_{2}}\left( {{t}_{1}}+{{t}_{2}} \right)}{2},\dfrac{a\left( {{t}_{1}}+3{{t}_{2}} \right)}{2} \right]...................\left( 3 \right) \\\ \end{aligned}
Now by using equation of line formula y−y1x−x1=y1−y2x1−x2\dfrac{y-{{y}_{1}}}{x-{{x}_{1}}}=\dfrac{{{y}_{1}}-{{y}_{2}}}{{{x}_{1}}-{{x}_{2}}}
⇒y−y1=(y1−y2x1−x2)(x−x1)................(4)\Rightarrow y-{{y}_{1}}=\left( \dfrac{{{y}_{1}}-{{y}_{2}}}{{{x}_{1}}-{{x}_{2}}} \right)\left( x-{{x}_{1}} \right)................\left( 4 \right)
Now substituting the values of (2) and (3) in (4)\left( 2 \right)\text{ and }\left( 3 \right)\text{ in }\left( 4 \right)
y−[a(3t1+t2)2]=(a(3t1+t2)2−a(t1+3t2)2at1(t1+t2)2−at2(t1+t2)2)[x−at1(t1+t2)2]y-\left[ \dfrac{a\left( 3{{t}_{1}}+{{t}_{2}} \right)}{2} \right]=\left( \dfrac{\dfrac{a\left( 3{{t}_{1}}+{{t}_{2}} \right)}{2}-\dfrac{a\left( {{t}_{1}}+3{{t}_{2}} \right)}{2}}{\dfrac{a{{t}_{1}}\left( {{t}_{1}}+{{t}_{2}} \right)}{2}-\dfrac{a{{t}_{2}}\left( {{t}_{1}}+{{t}_{2}} \right)}{2}} \right)\left[ x-\dfrac{a{{t}_{1}}\left( {{t}_{1}}+{{t}_{2}} \right)}{2} \right]
Simplifying and cancelling out like terms:
y-\dfrac{a\left( 3{{t}_{1}}+{{t}_{2}} \right)}{2}=\left( \dfrac{3a{{t}_{1}}+a{{t}_{2}}-a{{t}_{1}}-3a{{t}_{2}}}{a{{t}_{1}}^{2}+a{{t}_{1}}{{t}_{2}}-a{{t}_{1}}{{t}_{2}}-a{{t}_{2}}^{2}} \right)\left\\{ x-\dfrac{a{{t}_{1}}\left( {{t}_{1}}+{{t}_{2}} \right)}{2} \right\\}
y-\dfrac{a\left( 3{{t}_{1}}+{{t}_{2}} \right)}{2}=\left( \dfrac{2a{{t}_{1}}-2a{{t}_{2}}}{a{{t}_{1}}^{2}-a{{t}_{2}}^{2}} \right)\left\\{ x-\dfrac{a{{t}_{1}}\left( {{t}_{1}}+{{t}_{2}} \right)}{2} \right\\}(cancel out like terms)

& \Rightarrow y-\dfrac{a\left( 3{{t}_{1}}+{{t}_{2}} \right)}{2}=\left( \dfrac{1}{{{t}_{1}}+{{t}_{2}}} \right)\left\\{ x-\dfrac{a{{t}_{1}}\left( {{t}_{1}}+{{t}_{2}} \right)}{2} \right\\} \\\ & \left[ \because {{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) \right] \\\ & y=\dfrac{a\left( 3{{t}_{1}}+{{t}_{2}} \right)}{2}+\dfrac{x}{{{t}_{1}}+{{t}_{2}}}-\dfrac{a{{t}_{1}}}{2} \\\ & y=\dfrac{3a{{t}_{1}}-a{{t}_{1}}}{2}+\dfrac{a{{t}_{2}}}{2}+\dfrac{x}{{{t}_{1}}+{{t}_{2}}}\Rightarrow y=a{{t}_{1}}+\dfrac{a{{t}_{2}}}{2}+\dfrac{x}{{{t}_{1}}+{{t}_{2}}} \\\ & y=\dfrac{a\left( 2{{t}_{1}}+{{t}_{2}} \right)}{2}+\dfrac{x}{{{t}_{1}}+{{t}_{2}}} \\\ \end{aligned}$$ Now this equation is of the form $\Rightarrow y=mx+c$ And this tangent is on the parabola. Note: (i) The ortho-centre of the triangle formed by the tangents is on directrix and we have proved that the lines joining the midpoints of the side of the triangle are tangents on parabola. Hence the ortho-centre of the triangle formed by joining the midpoint of the sides of the triangle lies on directrix of the parabola. (ii) How to find the coordinates of points of intersection of tangents $A\left( a{{t}_{1}}^{2},2a{{t}_{1}} \right),B\left( a{{t}_{2}}^{2},2a{{t}_{2}} \right)$ Equation of parabola$\Rightarrow {{y}^{2}}=4ax$ The equation of tangent at ${{t}_{1}}\Rightarrow y{{t}_{1}}=x+a{{t}_{1}}^{2}.................\left( 1 \right)$ Equation of tangent at ${{t}_{2}}\Rightarrow y{{t}_{2}}=x+a{{t}_{2}}^{2}.................\left( 2 \right)$ $\begin{aligned} & \left( 1 \right)-\left( 2 \right)\Rightarrow y\left( {{t}_{1}}-{{t}_{2}} \right)=a\left( {{t}_{1}}^{2}-{{t}_{2}}^{2} \right) \\\ & \Rightarrow y\left( {{t}_{1}}-{{t}_{2}} \right)=a\left( {{t}_{1}}-{{t}_{2}} \right)\left( {{t}_{1}}+{{t}_{2}} \right) \\\ & \therefore y=a\left( {{t}_{1}}+{{t}_{2}} \right) \\\ \end{aligned}$ From equation $\left( 1 \right)\Rightarrow y{{t}_{1}}=x+a{{t}_{1}}^{2}\Rightarrow a\left( {{t}_{1}}+{{t}_{2}} \right)=x+a{{t}_{1}}^{2}$ $a{{t}_{1}}^{2}+a{{t}_{1}}{{t}_{2}}=x+a{{t}_{1}}^{2}\Rightarrow x=a{{t}_{1}}{{t}_{2}}$ $\therefore $ Point of intersection $=\left( x,y \right)=\left( a{{t}_{1}}{{t}_{2}},a\left( {{t}_{1}}+{{t}_{2}} \right) \right)$