Let P(4,4 √3) be a point on the parabola y^2=4 a x and PQ be a focal chord of the parabola…

Question

Let $\mathrm{P}(4,4 \sqrt{3})$ be a point on the parabola $y^2=4 \mathrm{a} x$ and PQ be a focal chord of the parabola. If M and N are the foot of perpendiculars drawn from P and Q respectively on the directrix of the parabola, then the area of the quadrilateral PQMN is equal to :

Accepted Answer

$$\begin{aligned}
&\begin{aligned}
& (4,4 \sqrt{3}) \text { lies on } \mathrm{y}^2=4 \mathrm{ax} \
& \Rightarrow 48=4 \mathrm{a} .4 \
& \quad 4 \mathrm{a}=12 \
& \Rightarrow \mathrm{y}^2=12 \mathrm{x} \text { is equation of parabola }
\end{aligned}\
\&\text { Now, parameter of } P \text { is } t_1=\frac{2}{\sqrt{3}} \Rightarrow \text { Parameters of } Q
\end{aligned}$$

$$\begin{aligned}
&\text { is } \mathrm{t}_2=-\frac{\sqrt{3}}{2} \Rightarrow \mathrm{Q}\left(\frac{9}{4},-3 \sqrt{3}\right)\
&\text { Area of trapezium PQNM }\
&\begin{aligned}
& =\frac{1}{2} \mathrm{MN} \cdot(\mathrm{PM}+\mathrm{QN}) \
& =\frac{1}{2} \mathrm{MN} \cdot(\mathrm{PS}+\mathrm{QS}) \
& =\frac{1}{2} \mathrm{MN} \cdot \mathrm{PQ} \
& =\frac{1}{2} 7 \sqrt{3} \cdot \frac{49}{4}=(343) \frac{\sqrt{3}}{8}=3
\end{aligned}
\end{aligned}$$

Solutions · 1