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If the points of intersection of the parabolas $y^2=5 x$ and $x^2=5 y$ lie on the line L, then the area of the triangle formed by the directrix of one parabola, latus rectum of another parabola and the line $\mathrm{L}$ is
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Verified Answer
The correct answer is:
$\frac{25}{8}$
$y^2=5 x$...(i)
$x^2=5 y$...(ii)
Solving (i) \& (ii)
$$
\begin{aligned}
& \left(\frac{x^2}{5}\right)^2=5 x \\
& x^4=x .5^3 \\
& \therefore x=0,5 \Rightarrow y=0,5
\end{aligned}
$$
Point of intersection $(0,0),(5,5)$
$\therefore \mathrm{L} \equiv y=x$...(1)
Equation of directix of $\mathrm{P}_1: x=\frac{-5}{4}$ Equation of latus rectum of $\mathrm{P}_2$ :
$$
\begin{aligned}
& y=\frac{5}{4} \\
& \mathrm{~A}\left(\frac{5}{4}, \frac{5}{4}\right), \mathrm{B}\left(\frac{-5}{4}, \frac{-5}{4}\right), \mathrm{C}\left(\frac{-5}{4}, \frac{5}{4}\right) \\
& \mathrm{Ar}(\triangle \mathrm{ABC})=\frac{1}{2}\left|\begin{array}{ccc}
\frac{5}{4} & \frac{5}{4} & 1 \\
\frac{-5}{4} & \frac{-5}{4} & 1 \\
\frac{-5}{4} & \frac{5}{4} & 1
\end{array}\right|=\frac{1}{2} \frac{5}{2} \times \frac{5}{2}=\frac{25}{8}
\end{aligned}
$$
$x^2=5 y$...(ii)
Solving (i) \& (ii)
$$
\begin{aligned}
& \left(\frac{x^2}{5}\right)^2=5 x \\
& x^4=x .5^3 \\
& \therefore x=0,5 \Rightarrow y=0,5
\end{aligned}
$$
Point of intersection $(0,0),(5,5)$
$\therefore \mathrm{L} \equiv y=x$...(1)
Equation of directix of $\mathrm{P}_1: x=\frac{-5}{4}$ Equation of latus rectum of $\mathrm{P}_2$ :
$$
\begin{aligned}
& y=\frac{5}{4} \\
& \mathrm{~A}\left(\frac{5}{4}, \frac{5}{4}\right), \mathrm{B}\left(\frac{-5}{4}, \frac{-5}{4}\right), \mathrm{C}\left(\frac{-5}{4}, \frac{5}{4}\right) \\
& \mathrm{Ar}(\triangle \mathrm{ABC})=\frac{1}{2}\left|\begin{array}{ccc}
\frac{5}{4} & \frac{5}{4} & 1 \\
\frac{-5}{4} & \frac{-5}{4} & 1 \\
\frac{-5}{4} & \frac{5}{4} & 1
\end{array}\right|=\frac{1}{2} \frac{5}{2} \times \frac{5}{2}=\frac{25}{8}
\end{aligned}
$$
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