SATHEE: Application of Derivatives 4 Question 44

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 $P Q \geq min (P P_{1}, Q Q_{1})$ . 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 $(t, t^{2} + 1)$

Reflection of $P$ in $y = x$ is $P_{1} (t^{2} + 1, t)$

which clearly lies on $y^{2} = x - 1$

Similarly, let coordinates of $Q$ be $(s^{2} + 1, s)$

Its reflection in $y = x$ is

$Q_{1} (s, s^{2} + 1)$ , which lies on $x^{2} = y - 1$ .

We have, $P Q_{1}^{2} = (t - s)^{2} + {(t^{2} - s^{2})}^{2} = P_{1} Q^{2}$

$\Rightarrow P Q_{1} = P_{1} Q$

Also $P P_{1} | Q Q_{1} [∵$ 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

$Missing or unrecognized delimiter for \left$

Let us take $Missing or unrecognized delimiter for \left$

$∴ P Q^{2} \geq P P_{1}^{2} = {(t^{2} + 1 - t)}^{2} + (t^{2} + 1 - t^{2})$

$= 2 (t^{2} + 1 - t^{2}) = f (t)$

[say] we have, $f^{'} (t) = 4 (t^{2} + 1 - t) (2 t - 1)$

$= 4 [(t - 1 / 2)^{2} + 3 / 4] [2 t - 1]$

Now, $f^{'} (t) = 0$

$\Rightarrow t = 1 / 2$

Also, $f^{'} (t) < 0$ for $t < 1 / 2$

and $f^{'} (t) > 0$ 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 $C_{2}$ .

Application of Derivatives 4 Question 44

Answer:

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Application of Derivatives 4 Question 44

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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