Hint: We start solving the problem by using the facts that the foot of the perpendicular from focus to any tangent of parabola lies on the tangent at the vertex of that parabola and it is the midpoint of the line segment joining the focus and the point on the directrix. We assume and find the point on the directrix using the midpoint formula $\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$. We then find the equation of the directrix with the point we obtained and also the point is given in the problem. We then find the axis using the property that the axis is perpendicular to the directrix of the parabola. We then find the intersection of the directrix and axis of parabola using which we find the vertex. We then find the tangent at the vertex using the vertex we obtained and the point given in the problem. We then the length of the latus rectum and perpendicular distance from focus latus rectum using the facts that they are four times and two times the distance between focus and vertex.
Complete step-by-step solution
According to the problem, we are given that the point $T\left( 3,2 \right)$ is the foot of the perpendicular drawn from the focus $S\left( 2,-1 \right)$ on a tangent and directrix passes through $P\left( 0,9 \right)$. We need to find which of the given options are true.
We know that the foot of the perpendicular from focus to any tangent of parabola lies on the tangent at the vertex of that parabola and it is the midpoint of the line segment joining the focus and the point on the directrix. Let us assume that the point on the directrix be $Q\left( a,b \right)$.
Let us draw all this information to get a better view.
We have got that the point $T\left( 3,2 \right)$ is the midpoint of the focus $S\left( 2,-1 \right)$ and the point $Q\left( a,b \right)$.
So, we have got $\left( \dfrac{2+a}{2},\dfrac{-1+b}{2} \right)=\left( 3,2 \right)$.
$\Rightarrow \dfrac{2+a}{2}=3$ and $\dfrac{-1+b}{2}=2$.
$\Rightarrow 2+a=6$ and $-1+b=4$.
$\Rightarrow a=4$ and $b=5$.
So, the point on the directrix is $Q\left( 4,5 \right)$.
Let us find the equation of the directrix which is passing through the points $P\left( 0,9 \right)$ and $Q\left( 4,5 \right)$.
We know that the equation of the line passing through the points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ is $y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\times \left( x-{{x}_{1}} \right)$.
So, the equation of the directrix is $y-9=\dfrac{5-9}{4-0}\times \left( x-0 \right)$.
$\Rightarrow y-9=\dfrac{-4}{4}\times x$.
$\Rightarrow y-9=-x$.
$\Rightarrow y=-x+9$ ---(1).
We know that if the equation of the line is represented as $y=mx+c$, then m is the slope of that line. Using this we get the slope of the directrix as -1.
We know that the axis of the parabola is perpendicular to the directrix of the parabola. We know that the product of the slopes of the two perpendicular lines is -1. Let us assume the slope of the axis be ‘m’.
So, we get $m\times -1=-1$.
$\Rightarrow m=1$.
We know that the axis of the parabola passes through its focus. So, we need to the equation of the passes through the point $S\left( 2,-1 \right)$ and having slope 1.
So, the equation of the axis is $y-\left( -1 \right)=1\left( x-2 \right)$.
$\Rightarrow y+1=x-2$.
$\Rightarrow x-y-3=0$.
So, we have found the equation of the axis of the parabola is $x-y-3=0$ ---(2).
Let us find the intersection point of the directrix and axis of the parabola.
Let us substitute equation (1) in equation (2).
So, we have $x-\left( -x+9 \right)-3=0$.
$\Rightarrow x+x-9-3=0$.
$\Rightarrow 2x-12=0$.
$\Rightarrow x=6$. Let us substitute this in equation (2).
So, we get $6-y-3=0$.
$\Rightarrow y=3$.
We have found the intersection point of the axis and the directrix of a hyperbola is $\left( 6,3 \right)$.
We know that the vertex of the parabola is the midpoint of the focus and intersection point of the axis and directrix.
Let us assume $\left( p,q \right)$ be the vertex of parabola.
So, we have $\left( p,q \right)=\left( \dfrac{6+2}{2},\dfrac{3-1}{2} \right)$.
$\Rightarrow \left( p,q \right)=\left( \dfrac{8}{2},\dfrac{2}{2} \right)$.
$\Rightarrow \left( p,q \right)=\left( 4,1 \right)$.
Now, we have the tangent at the vertex passing through the points $\left( 4,1 \right)$ and $T\left( 3,2 \right)$.
So, the equation of the tangent at vertex is $y-1=\dfrac{2-1}{3-4}\times \left( x-4 \right)$.
$\Rightarrow y-1=\dfrac{1}{-1}\times \left( x-4 \right)$.
$\Rightarrow y-1=-x+4$.
$\Rightarrow x+y-5=0$.
So, we have found the equation of the tangent at the vertex as $x+y-5=0$ ---(3).
Now, we know that the length of the latus rectum is four times the distance between focus and vertex.
So, let us find the distance between the vertex $\left( 4,1 \right)$ and focus $S\left( 2,-1 \right)$. Let us assume the distance is ‘r’.
$\Rightarrow r=\sqrt{{{\left( 4-2 \right)}^{2}}+{{\left( 1+1 \right)}^{2}}}$.
$\Rightarrow r=\sqrt{{{2}^{2}}+{{2}^{2}}}$.
$\Rightarrow r=\sqrt{4+4}$.
$\Rightarrow r=\sqrt{8}$.
$\Rightarrow r=2\sqrt{2}$ ---(4).
So, the length of the latus rectum = $4\times 2\sqrt{2}=8\sqrt{2}$ ---(5).
We know that the perpendicular distance from the focus to the directrix of the parabola is two times the distance between focus and vertex.
So, the perpendicular distance from focus to the directrix = $2\times 2\sqrt{2}=4\sqrt{2}$ ---(6).
From equations (2), (3), (5), and (6), we can see that the options (a), (b), (c) are correct.
$\therefore$ The correct options for the given problem are (a), (b), and (c).
Note: We can see that the given problem contains a huge amount of calculations which may lead us to confusion and making mistakes. We can also use the fact that the tangent at the vertex is parallel to the directrix of the parabola. We can also find the equation of the parabola using the vertex and equation of the axis we just obtained. We can also prove the facts that we used to find the point on the directrix by taking the tangent for the standard parabola which will be applicable for every parabola.
