Question

Let $P$ and $R$ are two points on the parabola $y=x^2$ such that $PS$ and $RS$ be the tangents at $P$ and $R,$ and $PQ$ and $RQ$ be the normal drawn at $P$ and $R$ respectively. If the length of rectangle $PQRS$ form is twice of its width, then which of the following is/are correct?

• A. The possible coordinates of $S$ are $(\frac{1}{4},\frac{1}{16})$ or $(-\frac{1}{4},\frac{1}{16})$
• B. The possible coordinates of $S$ are $(\frac{3}{8},-\frac{1}{4})$ or $(-\frac{3}{8},-\frac{1}{4})$
• C. Area of the rectangle $PQRS$ is $\frac{125}{128}$
• D. Length of the rectangle is $\frac{5\sqrt{5}}{8}$

Answer 1

Answer: (B), (C), (D)

The correct options are
B The possible coordinates of $S$ are $(\frac{3}{8},-\frac{1}{4})$ or $(-\frac{3}{8},-\frac{1}{4})$
C Area of the rectangle $PQRS$ is $\frac{125}{128}$
D Length of the rectangle is $\frac{5\sqrt{5}}{8}$

Let the coordinates of $P(p,p^2)$ and $R(r,r^2)$
$\therefore S=(\frac{p+r}{2},pr)$
Slope of the tangent,
$y=x^2\Rightarrow y^{\prime }=2x$
We know that the line $SP$ is perpendicular to $SR$, so
$2p\times 2r=-1\Rightarrow r=-\frac{1}{4p}\cdots \left(1\right)$
It is given that,
$SP=2SR\Rightarrow S{P}^2=4S{R}^2$
$\Rightarrow {\left(\frac{p-r}{2}\right)}^2+(p^2-pr{)}^2=4({\left(\frac{r-p}{2}\right)}^2+(r^2-pr{)}^2)$
$\Rightarrow -\frac{3}{4}(p-r{)}^2=(p-r{)}^2[4r^2-p^2]\Rightarrow p^2-4r^2=\frac{3}{4}[\because p\ne r]$
Using equation $\left(1\right)$,
$\Rightarrow p^2-\frac{4}{16p^2}=\frac{3}{4}\Rightarrow 4p^4-3p^2-1=0\Rightarrow (p^2-1)(4p^2+1)=0\Rightarrow p=\pm 1\Rightarrow r=\mp \frac{1}{4}$
Now, the coordinate of
$P=(-1,1)\text{or}(1,1)$
$R=(\frac{1}{4},\frac{1}{16})\text{or}(-\frac{1}{4},\frac{1}{16})S=(-\frac{3}{8},-\frac{1}{4})\text{or}(\frac{3}{8},-\frac{1}{4})$

Now, area of the rectangle $PQRS$
$=\left(PS\right)\times \left(SR\right)=\frac{(PS{)}^2}{2}=\frac{1}{2}[{\left(\frac{p-r}{2}\right)}^2+(p^2-pr{)}^2]=\frac{1}{2}\left[\right(p-r{)}^2(p^2+\frac{1}{4})]=\frac{1}{2}\times \frac{5^2}{4^2}\times \frac{5}{4}=\frac{125}{128}\therefore PS=\sqrt{\frac{125}{64}}=\frac{5\sqrt{5}}{8}$