Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Management
  • The lines <img alt="p\left ( p^{2}+1 \right )x-y+q= 0" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cdpi%7B100%7D%20p%5Cleft%20%28%20p%5E%7B2%7D&plus;1%20%5Cright%20%29x-y&plus;q%

The lines p\left ( p^{2}+1 \right )x-y+q= 0 and

\left ( p^{2}+1 \right )^{2}\! x+\left ( p^{2}+1 \right )y+2q= 0 are perpendicular to a common line for

Option: 1

exactly one value of  p


Option: 2

exactly two value of  p


Option: 3

more than two value of  p


Option: 4

No value of  p


Answers (1)

best_answer

If two lines are perpendicular to same line, they must be parallel.

p(p^{2}+1)x -y+q=0

(p^{2}+1)^{2}x + (p^{2}+1)y+2q=0

Slope of line 1 = slope of line 2

-\frac{p(p^{2}+1)}{-1}=-\frac{(p^{2}+1)^{2}}{(p^{2}+1)}\Rightarrow p=-1

So, we have only one value of p

 

Posted by

rishi.raj

View full answer

Similar Questions

Trending Articles/News

anam-khan

CAT 2024: In IIM Shillong's MBA admission criteria, 40% weightage to exam
anam-khan

CAT Exam 2024 Question Paper: How much time should you spend on DILR, VARC, Quant?
anam-khan

CAT 2024: IIM Bangalore gives 25% weightage to exam for MBA admissions

Latest Question