Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
0 votes

The shortest distance between the line $y-x=1$ and the curve $x=y^2$ is

$\begin{array}{1 1}(A)\;\frac{2 \sqrt 3}{8} \\(B)\;\frac{3 \sqrt 2}{5} \\(C)\;\frac{\sqrt 3}{4} \\(D)\; \frac {3 \sqrt 2}{8} \end{array}$

Can you answer this question?

1 Answer

0 votes
Let $(a^2,a)$ be thw point of shortest distance on $x=y^2$
The distance between $(a^2,a)$ and line $x-y+1=0$ is given by
$D= \large\frac{a^2-a+1}{\sqrt 2}$
$\qquad= \large\frac{1}{\sqrt 2} \bigg[ (a- \large\frac{1}{2})^2+\frac{3}{4}\bigg]$
If is min when $a= \large\frac{1}{2} $ and $D_{min}=\large\frac{3}{4 \sqrt 2}$
$=> \large\frac{3 \sqrt 2}{8}$
Hence D is the correct answer.
answered Apr 8, 2014 by meena.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App