A circle is described whose centre is the vertex and whose diamet… - Vidyadip

MCQ

A circle is described whose centre is the vertex and whose diameter is three-quarters of the latus rectum of the parabola $ y^2 = 4ax.$ If $PQ$ is the common chord of the circle and the parabola and $ L_1 L_2$ is the latus rectum, then the area of the trapezium $PL_1 L_2Q$ is :

A
$3\ \sqrt 2\ a^2$
B
$\left( {\frac{{\sqrt 2 \, + 1}}{2}} \right)\,{a^2}\,$
C
$4 a^2$
✓
$\left( {\frac{{2\,\, + \,\,\sqrt 2 }}{2}} \right) a^2$

✓

Answer

Correct option: D.

$\left( {\frac{{2\,\, + \,\,\sqrt 2 }}{2}} \right) a^2$

d
Centre $(0,0),$ radius $=\frac{1}{2} \cdot \frac{3}{4} \cdot 4 a=\frac{3 a}{2}$

Equation of the circle is $4\left(x^{2}+y^{2}\right)=9 a^{2} \ldots \ldots \ldots$ (i) Equation of the parabola is $y^{2}=4 a x$ $\ldots \ldots$

(ii)

Solving (i) and

(ii) $x^{2}+4 a x-\frac{9 a^{2}}{4}$

$x=\frac{-4 a \pm \sqrt{16 a^{2}+9 a^{2}}}{2}=\frac{-4 a \pm 5 a}{2}$

$\therefore \quad x=a / 2$

$x=a / 2 \quad y^{2}=4 a x=4 a a / 2=2 a^{2}$

$\Rightarrow \quad y=\pm \sqrt{2} a$

$\therefore=2 \sqrt{2} a$

$\therefore \quad=\frac{1}{2}\left(a-\frac{a}{2}\right)(4 a+2 \sqrt{2} \quad a)=\left(\frac{2+\sqrt{2}}{2}\right) a^{2}$

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