If the points of intersection of the parabola | vedclass

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(C) Given parabolas are $P_1: y^2=5x$ and $P_2: x^2=5y$.Solving these equations: Substituting $y = \frac{x^2}{5}$ into $y^2=5x$ gives $(\frac{x^2}{5})

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$\frac{25}{8}$

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(C) Given parabolas are $P_1: y^2=5x$ and $P_2: x^2=5y$.Solving these equations: Substituting $y = \frac{x^2}{5}$ into $y^2=5x$ gives $(\frac{x^2}{5})^2 = 5x$,which simplifies to $x^4 = 125x$. Thus,$x(x^3 - 125) = 0$,giving $x=0$ or $x=5$.The points of intersection are $(0,0)$ and $(5,5)$.The line $L$ passing through $(0,0)$ and $(5,5)$ is $y=x$.The directrix of $P_1$ ($y^2=4ax$ with $4a=5 \Rightarrow a=5/4$) is $x = -\frac{5}{4}$.The latus rectum of $P_2$ ($x^2=4ay$ with $4a=5 \Rightarrow a=5/4$) is $y = \frac{5}{4}$.The vertices of the triangle are the intersection points of these three lines:$1$. Intersection of $x = -\frac{5}{4}$ and $y = \frac{5}{4}$ is $C(-\frac{5}{4}, \frac{5}{4})$.$2$. Intersection of $x = -\frac{5}{4}$ and $y=x$ is $B(-\frac{5}{4}, -\frac{5}{4})$.$3$. Intersection of $y = \frac{5}{4}$ and $y=x$ is $A(\frac{5}{4}, \frac{5}{4})$.The area of the triangle with vertices $A(\frac{5}{4}, \frac{5}{4})$,$B(-\frac{5}{4}, -\frac{5}{4})$,and $C(-\frac{5}{4}, \frac{5}{4})$ is given by:Area $= \frac{1}{2} |x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)|$Area $= \frac{1}{2} |\frac{5}{4}(-\frac{5}{4} - \frac{5}{4}) + (-\frac{5}{4})(\frac{5}{4} - \frac{5}{4}) + (-\frac{5}{4})(\frac{5}{4} - (-\frac{5}{4}))|$Area $= \frac{1}{2} |\frac{5}{4}(-\frac{10}{4}) + 0 + (-\frac{5}{4})(\frac{10}{4})| = \frac{1}{2} |-\frac{50}{16} - \frac{50}{16}| = \frac{1}{2} |-\frac{100}{16}| = \frac{50}{16} = \frac{25}{8}$.

If the points of intersection of the parabolas $y^2=5x$ and $x^2=5y$ lie on the line $L$,then the area of the triangle formed by the directrix of one parabola,the latus rectum of another parabola,and the line $L$ is

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