Given the parabola $x^2=-4 y$, a point $\mathrm{P}(4,-4)$ lies on it. A focal chord PQ is drawn thro

Question

Given the parabola $x^2=-4 y$, a point $\mathrm{P}(4,-4)$ lies on it. A focal chord PQ is drawn through P . Let M and N be the points where perpendiculars from P and Q intersect the directrix of the parabola. Find the area of the quadrilateral PQMN .

Professor PC Thomas Classes · Answer

The given parabola is $\boldsymbol{x}^2=-\mathbf{4} \boldsymbol{y}$. Comparing this with the standard form $\boldsymbol{x}^2=-\mathbf{4} \boldsymbol{a} \boldsymbol{y}$, we find $\boldsymbol{a}=\mathbf{1}$.
The focus of the parabola is $\mathbf{F}(\mathbf{0},-\boldsymbol{a})=(\mathbf{0},-\mathbf{1})$.
The equation of the directrix is $\boldsymbol{y}=\boldsymbol{a}$, so $\boldsymbol{y}=\mathbf{1}$.
Let $\mathbf{P}\left(x_1, y_1ight)=(4,-4)$.
Since $\mathbf{M}$ is the foot of the perpendicular from $\mathbf{P}$ to the directrix $\boldsymbol{y}=\mathbf{1}$, the coordinates of $\mathbf{M}$ are $\left(\boldsymbol{x}_1, \mathbf{1}ight)=(4,1)$.
For a parabola $\boldsymbol{x}^{\mathbf{2}}=-\boldsymbol{4} \boldsymbol{a} \boldsymbol{y}$, a point on the parabola can be represented parametrically as $\left(\mathbf{2} \boldsymbol{a} \boldsymbol{t},-\boldsymbol{a} \boldsymbol{t}^{\mathbf{2}}ight)$.
For $\mathbf{P}(4,-4)$ and $\boldsymbol{a}=1$, we have $2 t=4 \Rightarrow t=2$. Also, $-t^2=-4$, which gives $t^2=4$, so $t= \pm 2$. Thus, $t=2$ for point $\mathbf{P}$.
Since $\mathbf{P Q}$ is a focal chord, if $\mathbf{P}$ corresponds to parameter $\boldsymbol{t}$, then $\mathbf{Q}$ corresponds to parameter $-1 / t$.
So, for Q , the parameter is $t^{\prime}=-1 / 2$.
The coordinates of Q are $\left(2 a(-1 / t),-a(-1 / t)^2ight)=\left(2(1)(-1 / 2),-(1)(-1 / 2)^2ight)=(-1,-1 / 4)$.
Since $\mathbf{N}$ is the foot of the perpendicular from $\mathbf{Q}$ to the directrix $\boldsymbol{y}=\mathbf{1}$, the coordinates of $\mathbf{N}$ are $\left(\boldsymbol{x}_Q, \mathbf{1}ight)=(-\mathbf{1}, \mathbf{1})$.
Now we have the vertices of the quadrilateral PQMN:
$$
\begin{aligned}
& \mathrm{P}(4,-4) \
& \mathrm{Q}(-1,-1 / 4) \
& \mathrm{M}(4,1) \
& \mathrm{N}(-1,1)
\end{aligned}
$$ Notice that PM and QN are vertical line segments (both perpendicular to the directrix $\boldsymbol{y}=\mathbf{1}$ ). Thus, PM is parallel to QN. This means PQMN is a trapezoid.
The lengths of the parallel sides are:
Length of $\mathrm{PM}=\left|y_P-y_Might|=|-4-1|=|-5|=5$.
Length of QN $=\left|y_Q-y_Night|=|-1 / 4-1|=|-5 / 4|=5 / 4$.
The height of the trapezoid is the perpendicular distance between the lines $\boldsymbol{x}=\mathbf{4}$ (containing PM ) and $\boldsymbol{x}=-\mathbf{1}$ (containing QN ), which is the difference in their $x$-coordinates: Height $=\left|x_P-x_Qight|=|4-(-1)|=|5|=5$.
The area of a trapezoid is given by $\frac{1}{2} 	imes$ (sum of parallel sides) × (height).
Area of $\mathrm{PQMN}=\frac{1}{2} 	imes(\mathrm{PM}+\mathrm{QN}) 	imes($ distance between PM and QN$)$
Area $=\frac{1}{2} 	imes\left(5+\frac{5}{4}ight) 	imes 5$
Area $=\frac{1}{2} 	imes\left(\frac{20+5}{4}ight) 	imes 5$
Area $=\frac{1}{2} 	imes \frac{25}{4} 	imes 5$
Area $=\frac{125}{8}$.
Thus, the area of the quadrilateral PQMN is $\frac{125}{8}$.