Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

Find curve in $x-y$ plane which passes through$(1,1)$ . Intersects line $y=x$ at right angle at that point and satisfies $xy''+2y''=0$

$(a)\;y=\frac{1}{x^2} \\ (b)\;y = \frac{1}{x} \\ (c)\;y=0 \\ (d)\;y=x^2 $
Can you answer this question?

1 Answer

0 votes
$y'= t$
$x \large\frac{dt}{dx}$$=-2t$
$\large\frac{dt}{t}=\frac{-2 dx}{x}$
$\log t = -2 \log x +\log c$
$t= \large\frac{c}{x^2}$
$dy =\large\frac{c dx}{x^2}$
$y= \large\frac{-c}{x}$$+c_1$
It satisfies $(1,1)$
Curve intersects the line $y=x$ at right angle means it is normal to curve.
So, $\large\frac{dy}{dx}=\frac{c}{x^2}$$=-1$
So, $c_1=0$
So, $y= \large\frac{1}{x}$
Hence b is the correct answer.
answered Feb 6, 2014 by meena.p

Related questions

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