Application of Derivatives 4 Question 44
46. Let $C _1$ and $C _2$ be respectively, the parabolas $x^{2}=y-1$ and $y^{2}=x-1$. Let $P$ be any point on $C _1$ and $Q$ be any point on $C _2$. If $P _1$ and $Q _1$ is the reflections of $P$ and $Q$, respectively, with respect to the line $y=x$. Prove that $P _1$ lies on $C _2 Q _1$ lies on $C _1$ and $P Q \geq \min \left(P P _1, Q Q _1\right)$. Hence, determine points $P _0$ and $Q _0$ on the parabolas $C _1$ and $C _2$ respectively such that $P _0 Q _0 \leq P Q$ for all pairs of points $(P, Q)$ with $P$ on $C _1$ and $Q$ on $C _2$.
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Answer:
Correct Answer: 46. $b \in(-2,-1) \cup[1, \infty]$
Solution:
- Let coordinates of $P$ be $\left(t, t^{2}+1\right)$
Reflection of $P$ in $y=x$ is $P _1\left(t^{2}+1, t\right)$
which clearly lies on $y^{2}=x-1$
Similarly, let coordinates of $Q$ be $\left(s^{2}+1, s\right)$
Its reflection in $y=x$ is
$Q _1\left(s, s^{2}+1\right)$, which lies on $x^{2}=y-1$.
We have, $\quad P Q _1^{2}=(t-s)^{2}+\left(t^{2}-s^{2}\right)^{2}=P _1 Q^{2}$
$\Rightarrow \quad P Q _1=P _1 Q$
Also $P P _1 | Q Q _1 \quad[\because$ both perpendicular to $y=x]$
Thus, $P P _1 Q Q _1$ is an isosceles trapezium.
Also, $P$ lies on $P Q _1$ and $Q$ lies on $P _1 Q$, then
$P Q \geq \min {P P _1 Q Q _1 }$
Let us take $\min {P P _1 Q Q _1 }=P P _1$
$\therefore \quad P Q^{2} \geq P P _1^{2}=\left(t^{2}+1-t\right)^{2}+\left(t^{2}+1-t^{2}\right)$
$$ =2\left(t^{2}+1-t^{2}\right)=f(t) $$
[say] we have, $f^{\prime}(t)=4\left(t^{2}+1-t\right)(2 t-1)$
$$ =4\left[(t-1 / 2)^{2}+3 / 4\right][2 t-1] $$
Now, $\quad f^{\prime}(t)=0$
$\Rightarrow \quad t=1 / 2$
Also, $f^{\prime}(t)<0$ for $t<1 / 2$
and $f^{\prime}(t)>0$ for $t>1 / 2$
Thus, $f(t)$ is least when $t=1 / 2$.
Corresponding to $t=1 / 2$, point $P _0$ on $C _1$ is $(1 / 2,5 / 4)$ and $P _1$ (which we take as $Q _0$ ) on $C _2$ are $(5 / 4,1 / 2)$. Note that $P _0 Q _0 \leq P Q$ for all pairs of $(P, Q)$ with $P$ on $C _2$.
