Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The shortest distance between the line    and the curve  <img alt="\m

The shortest distance between the line  \mathrm { y=x }  and the curve  \mathrm { y^2=x-2 \: \: is \: \frac{a}{b \sqrt{2}} }   units. Then value of \mathrm { a+b }  is

Option: 1

12


Option: 2

13


Option: 3

11


Option: 4

10


Answers (1)

best_answer

We are given the line  \mathrm {y=x \Rightarrow \frac{d y}{d x}=1}  and the curve  \mathrm {y^2=x-2}

\begin{aligned} & \Rightarrow \mathrm {2 y \frac{d y}{d x}=1} \\ & \Rightarrow \mathrm {\frac{d y}{d x}=\frac{1}{2 y}} \end{aligned}

Let   \mathrm {P\left(t^2+2, t\right)}  be any point on the curve.
Thus, \mathrm {\left(\frac{d y}{d x}\right)_{\left(t^2+2, t\right)}=\frac{1}{2 t}} 
Now, \mathrm {\frac{1}{2 t}=1 \Rightarrow 2 t=1 \Rightarrow t=\frac{1}{2}. \: \: So, P \equiv\left(\frac{9}{4}, \frac{1}{2}\right)}

Hence, shortest distance

\begin{aligned} &=\frac{\left|\frac{9}{4}-\frac{1}{2}\right|}{\sqrt{1^2+1^2}}=\frac{7}{4 \sqrt{2}} \text { units } \\ & =\mathrm {\frac{a}{b \sqrt{2}} \text { units } }\\ & \Rightarrow \mathrm {a=7, b=4, a+b=11} \end{aligned}

Posted by

vishal kumar

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Latest Question