A point on the parabola whose focus is $S(1,-1)$ and whose vertex is $A(1,1)$ is
A point on the parabola whose focus is $S(1,-1)$ and whose vertex is $A(1,1)$ is
$\left(3, \frac{1}{2}\right)$
$(1,2)$
$\left(2, \frac{1}{2}\right)$
$(2,2)$
Solution
(a) The gradient of the line joining the focus $S(1,-1)$ and vertex $A(1,1)$ is
$
m=\frac{-1-1}{1-1}=0
$
Let $Q(h, k)$ be the point of intersection of the axis $A S$ with the directrix. The $A(1,1)$ will be the mid-point of $Q S$.
$
\begin{array}{rlrl}
& \therefore & & h+1 \\
2 & & 1 \text { and } \frac{k-1}{2}=1 \\
& \Rightarrow & h & =1 \text { and } k=3
\end{array}
$
$\therefore Q$ is the point $(1,3)$
So, the directrix passes through the point $(1,3)$ and has the gradient 0 .
The equation of the directrix is
$
y-3=0
$
Let $P(x, y)$ be any point on the parabola and $M$ be the foot of the perpendicular drawn from $P$ on the directrix.
$\therefore$
As,
$
\begin{aligned}
P S & =P M \\
P S^2 & =P M^2
\end{aligned}
$
As, $(x-1)^2+(y+1)^2=\left(\frac{y-3}{\sqrt{1}}\right)^2$
$
\begin{array}{rlrl}
\Rightarrow & (x-1)^2+(y+1)^2 & =(y-3)^2 \\
\Rightarrow & & (x-1)^2 & =8(1-y)
\end{array}
$
By checking option (a),
$
\begin{aligned}
\Rightarrow & & (3-1)^2 & =8\left(1-\frac{1}{2}\right) \\
\Rightarrow & & (2)^2 & =8 \times \frac{1}{2} \\
\Rightarrow & & 4 & =4
\end{aligned}
$
Hence, point $\left(3, \frac{1}{2}\right)$ lies on the parabola
$
(x-1)^2=8(1-y) .
$