Thank you for reporting, we will resolve it shortly
Q.
The shortest distance between the line $y-x=1$ and the curve $x = y^2$ is
AIEEEAIEEE 2009Straight Lines
Solution:
$x − y + 1= 0 \dots\left(1\right)$
$x=y^{2}$
$1-2y \frac{dy}{dx} \Rightarrow \frac{dy}{dx}=\frac{1}{2y}=$ Slope of given line $\left(1\right)$
$\frac{1}{2y}=1 \Rightarrow y=\frac{1}{2} \Rightarrow x=\left(\frac{1}{2}\right)^{2}=\frac{1}{4} \Rightarrow \left(x, y\right)=\cdot\left(\frac{1}{4}, \frac{1}{2}\right)$
$\therefore $ The shortest distance is $\frac{\left|\frac{1}{4}-\frac{1}{2}+1\right|}{\sqrt{1+1}}=\frac{3}{4\sqrt{2}}=\frac{3\sqrt{2}}{8}$ Directions: Question number 86 to 90 are Assertion - Reason type questions. Each of these questions contains two statements Statement-1 (Assertion) and Statement-2 (Reason).
Each of these questions also have four alternative choices, only one of which is the correct answer. You have to select the correct choice