Question

Prove that the equation of the director circle to the parabola \[{{y}^{2}}=4ax\] is x + a = 0.

Answer 1

Hint: The director circle of a parabola is the directrix of the curve. First draw  the point of intersection of the tangent to the parabola whose equation is \[{{y}^{2}}=4ax\]. Consider the end points of the focal part on the parabola as \[P\left( a{{t}^{2}},2at \right)\] and \[{{P}^{'}}\left( at_{1}^{2},2a{{t}_{1}} \right)\]. Start with differentiating the equation of parabola. 

Complete step-by-step answer:

We know that the equation of parabola is \[{{y}^{2}}=4ax.....\left( i \right)\]

The figure of the parabola given below illustrates the focus, directrix, axis and vertex of the parabola.

We know that the tangents at the extremities of any focal chord intersect at right angles on the directrix. Let us take the end points of the focal part on the parabola as \[P\left( a{{t}^{2}},2at \right)\] and \[{{P}^{'}}\left( at_{1}^{2},2a{{t}_{1}} \right)\].

To find the equation of tangent to parabola at a given point, we need to differentiate the equation of parabola,

\[{{y}^{2}}=4ax\]

Differentiating both sides,

$ 2y=4a $

$  \dfrac{dy}{dx}=\dfrac{2a}{y} $

Therefore, the slope of a tangent at a given point \[\left( {{x}_{1}},{{y}_{1}} \right)\] on the parabola is:

\[{{m}_{t}}={{\left[ \dfrac{dy}{dx} \right]}_{\left( {{x}_{1}},{{y}_{1}}\right)}}=\dfrac{2a}{{{y}_{1}}}\]

Equation of line passing through \[\left( {{x}_{1}},{{y}_{1}} \right)\] having slope m by point – slope form

\[\Rightarrow y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)\]

Therefore, equation of tangent

\[y-{{y}_{1}}={{m}_{t}}\left( x-{{x}_{1}} \right)\]

\[y-{{y}_{1}}=\dfrac{2a}{{{y}_{1}}}\left( x-{{x}_{1}} \right)\]

\[{{y}_{1}}\left( y-{{y}_{1}} \right)=2ax-2a{{x}_{1}}\]

\[{{y}_{1}}y-y_{1}^{2}=2ax-2a{{x}_{1}}\]

\[{{y}_{1}}y-4a{{x}_{1}}=2ax-2a{{x}_{1}}\]

\[\Rightarrow {{y}_{1}}y=2ax+2a{{x}_{1}}\]

\[\therefore y{{y}_{1}}=2a\left( x+{{x}_{1}} \right)\]

At point \[P\left( at_{1}^{2},2at \right)\], the equation of tangent is

\[\Rightarrow ty=x+a{{t}^{2}}.....\left( ii \right)\]

At point \[{{P}^{'}}\left( at_{1}^{2},2a{{t}_{1}} \right)\], the equation of tangent is

\[\Rightarrow {{t}_{1}}y=x+at_{1}^{2}.....\left( iii \right)\]

Therefore, the slope \[t{{t}_{1}}=-1\]

Now, let three tangents intersect at a point \[\left( h,k \right)\].

The coordinates of points of intersection of tangents in a parabola can be found by subtracting the equation of tangents at the 2 points P and P’ respectively.

i.e. Equation (ii) – Equation (iii)

Subtracting both the equation \[ ({{a}^{2}}-{{b}^{2}})=\left( a-b \right)\left( a+b \right) \]

$  ty=x+a{{t}^{2}} $

$  \underline{{{t}_{1}}y=x+at_{1}^{2}\text{ }} $

$  y\left( t-{{t}_{1}} \right)=a\left( {{t}^{2}}-t_{1}^{2} \right) $

\[y\left( t-{{t}_{1}} \right)=a\left( t-{{t}_{1}} \right)\left( t+{{t}_{1}} \right)\]

\[\Rightarrow y=a\left( t+{{t}_{1}} \right)\]

Substitute the value of y in equation (ii).

\[ty=x+a{{t}^{2}}\]

\[\Rightarrow ta\left( t+{{t}_{1}} \right)=x+a{{t}^{2}}\]

\[a{{t}^{2}}+at{{t}_{1}}=x+a{{t}^{2}}\]

\[\Rightarrow x=at{{t}_{1}}\]

Putting the value of x to find y in equation (ii).

\[yt=x+a{{t}^{2}}\]

\[yt=at{{t}_{1}}+a{{t}^{2}}\]

\[\Rightarrow yt=at\left( {{t}_{1}}+t \right)\]

\[\Rightarrow y=a\left( t+{{t}_{1}} \right)\]

\[\therefore \left( x,y \right)=\left( at{{t}_{1}},a\left( t+{{t}_{1}} \right) \right)\]

(x, y) can be taken as (h, k)

\[\therefore h=at{{t}_{1}}\] and \[k=a\left( t+{{t}_{1}} \right)\].

Since, the tangents are perpendicular, \[t{{t}_{1}}=-1\]

\[\Rightarrow h=a\left( -1 \right)\]

\[\Rightarrow h=-a\]

\[\Rightarrow h+a=0\]

Where h = x.

\[\therefore x+a=0\]

Hence, the locus of the point (h, k) is x = -a which is the equation of the directrix.

Note: The chord joining \[\left( a{{t}^{2}},2at \right)\] and \[\left( at_{1}^{2},2a{{t}_{1}} \right)\] passes through focus, if \[{{t}_{1}}{{t}_{2}}=-1\]. This is formed because the tangents are perpendicular.