Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 1 - Precalculus Review - 1.2 Linear and Quadratic Functions - Exercises - Page 19: 54


The prove is as below.

Work Step by Step

Here, the point $P$ is on the parabola $y=x^2$, so, the coordinate will be $(x, x^2)$. Now, the distance between point $\left(0, \dfrac{1}{4}\right)$ and $P(x, x^2)$ can be calculated using distance formula as follows: Distance, $d_{1}=\sqrt{(x-0)^2+\left(x^2-\dfrac{1}{4}\right)^2}$ $\Rightarrow d_{1}=\sqrt{x^2+(x^2)^2-2(x^2)\left(\dfrac{1}{4}\right)+\left(\dfrac{1}{4}\right)^2}$ $\Rightarrow d_{1}=\sqrt{x^2+(x^2)^2-\dfrac{x^2}{2}+\left(\dfrac{1}{4}\right)^2}$ $\Rightarrow d_{1}=\sqrt{(x^2)^2+\dfrac{x^2}{2}+\left(\dfrac{1}{4}\right)^2}$ $\Rightarrow d_{1}=\sqrt{(x^2)^2+2(x^2)\left(\dfrac{1}{4}\right)+\left(\dfrac{1}{4}\right)^2}$ $\Rightarrow d_{1}=\sqrt{\left(x^2+\dfrac{1}{4}\right)^2}$ $\Rightarrow d_{1}=x^2+\dfrac{1}{4}$ units. Further, point $P(x, x^2)$ is $x^2$ units above x-axis and the line $y=-\dfrac{1}{4}$ is $\dfrac{1}{4}$ units below x-axis, so, the distance between point $P$ and line $y=-\dfrac{1}{4}$ is $d_{2}=x^2+\dfrac{1}{4}$ units. Therefore, we have $d_{1}=d_{2}=x^2+\dfrac{1}{4}$. Hence, proved.
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