Question

Let $AB$ be the latus rectum of the parabola $y^2=4ax$ in the $xy-plane$. Let $T$ be the region bounded by the finite arc $AB$ of the parabola and the lie segment $AB$. A rectangle $PQRS$ of maximum possible area is inscribed in $T$ with $P,Q$ on line $AB$m and $R,S$ on arc $AB$, Then $area\left(PQRS\right)/area\left(T\right)$ equals.

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{1}{\sqrt{3}}$

Answer 1

The correct option is D $\frac{1}{\sqrt{3}}$

$y^2=4ax$ and $AB$ is the latus rectum hence point $F(a,0)$

now $area\left(T\right)=$ area bound by the parabolic arc $AB$ and the latus rectum $AB$

now wee can write the parabola equation as $y=\pm 2\sqrt{ax}$

taking the positive side of the curve i.e the parabola above $x-axis$

hence $area\left(T\right)=2\times$area bound by $y=2\sqrt{ax}$from $x=0$ to $x=a$(since the parabola is symmetrical with respect to the x axis

$area\left(T\right)=2\times {\int }_0^{a}2\sqrt{ax}$

after simplyfying and putting the limits we get

$area\left(T\right)=\frac{8}{3}{a}^2$

since $PQRS$is a $rectangle$ and the points $P,Q$ lie on the latus rectum and are parallel to the y axis hence the line $RS$ is also parallel to yaxis and to the latus rectum .

any point on the parabola can be expressed as $S(a{t}^2,2at)$

by the symmetry of the situation we can deduce that point $S$ Is the image of point $R$ with respect to the x axis so the point $R(a{t}^2,-2at)$

hence the points $P(a,2at)$and $Q(a,-2at)$

now $PS=a-a{t}^2$(since $PS$ is parallel to the $x-axis$)

and $RS=4at$(since $RS$ is parallel to $y-axis$)

$area\left(PQRS\right)=PS\times RS$

$=4a^{2}t(1-t^2)$

given that we have to maximize the area of the rectangle i.e maximize the function$f\left(t\right)=4a^{2}t(1-t^2{)}^{=}4a^2(t-t^3)$

for the area to bee maximum $f(t{)}^{\prime }=0$

or $4a^2(1-3t^2)=0$

or $t=\pm \frac{1}{\sqrt{3}}$

taking t to be positive i.e $t=\frac{1}{\sqrt{3}}$and substituting this value of $t$in the $f\left(t\right)$

$area\left(PQRS\right)=4a^2\times \frac{1}{\sqrt{3}}\times (1-\frac{1}{3})$

or $area\left(PQRS\right)=\frac{8}{3\sqrt{3}}{a}^2$

hence the ratio $\frac{area\left(PQRS\right)}{area\left(T\right)}=\frac{\frac{8}{3}{a}^2}{\frac{8}{3\sqrt{3}}{a}^2}=\frac{1}{\sqrt{3}}$