Solveeit Logo
Login

Question

Question: The lines \[p({{p}^{2}}+1)x-y+q=0\]and \[{{({{p}^{2}}+1)}^{2}}x+({{p}^{2}}+1)y+2q=0\]are perpendicul...

The lines p(p2+1)x−y+q=0p({{p}^{2}}+1)x-y+q=0and (p2+1)2x+(p2+1)y+2q=0{{({{p}^{2}}+1)}^{2}}x+({{p}^{2}}+1)y+2q=0are perpendicular to common line for
A.No value of p
B.Exactly one value of p
C.Exactly two values of p
D.More than two values of p

Explanation

Solution

Hint : In this question it is written that two lines of equation are perpendicular to the common line that means two lines of equation are parallel to each other. Therefore, m1=m2{{m}_{1}}={{m}_{2}}is the condition applied when two lines of equation are parallel to each other. m1={{m}_{1}}=Slope of first line of equation. m2={{m}_{2}}=Slope of second line of equation.

Complete step-by-step answer :
Here, in this question it is given that there are two lines of equation.
That is
p(p2+1)x−y+q=0−−−−(1)p({{p}^{2}}+1)x-y+q=0----(1)
(p2+1)2x+(p2+1)y+2q=0−−−−(2){{({{p}^{2}}+1)}^{2}}x+({{p}^{2}}+1)y+2q=0----(2)
So we have to find the value of p.
For that we need to observe which is written in the question: The common line is perpendicular to equation (1) and equation (2). The meaning of this statement is that equation (1) and equation (2) are parallel to each other.
Hence the condition would be m1=m2{{m}_{1}}={{m}_{2}}
For that we have to find the value of m1{{m}_{1}}and m2{{m}_{2}}
We will get the value of m1{{m}_{1}} from equation (1)
p(p2+1)x−y+q=0p({{p}^{2}}+1)x-y+q=0
Equation (1) can be written in the form of y=mx+cy=mx+cthen, we get:
y=p(p2+1)x+qy=p({{p}^{2}}+1)x+q
Therefore, we get m1=p(p2+1){{m}_{1}}=p({{p}^{2}}+1) (By comparing with standard form)
We will get the value of m2{{m}_{2}} from equation (2)
(p2+1)2x+(p2+1)y+2q=0{{({{p}^{2}}+1)}^{2}}x+({{p}^{2}}+1)y+2q=0
Equation (1) can be written in the form of y=mx+cy=mx+cthen, we get:
y=−(p2+1)2(p2+1)x−2qy=\dfrac{-{{({{p}^{2}}+1)}^{2}}}{({{p}^{2}}+1)}x-2q
Therefore, we get m2=−(p2+1)2(p2+1){{m}_{2}}=\dfrac{-{{({{p}^{2}}+1)}^{2}}}{({{p}^{2}}+1)}
By simplifying this we get:
m2=−(p2+1){{m}_{2}}=-({{p}^{2}}+1)
Substitute the value of m1{{m}_{1}}and m2{{m}_{2}}on this condition m1=m2{{m}_{1}}={{m}_{2}}
p(p2+1)=−(p2+1)p({{p}^{2}}+1)=-({{p}^{2}}+1)
Here, (p2+1)({{p}^{2}}+1)get cancelled and we get:
p=−1p=-1
Hence, the lines are perpendicular to a common line for exactly one value of p.
So, the correct option is “option B”.
So, the correct answer is “Option B”.

Note : In this particular problem, you will make mistakes while understanding the question: the common line is perpendicular to equation (1) and equation (2). And you apply m1m2=−1{{m}_{1}}{{m}_{2}}=-1 that is wrong because in this case m2{{m}_{2}} is the slope of required line (common line) so, you can’t apply this. But if you observe this statement common line perpendicular to two line equation that means two lines of equations are parallel to each other hence you can apply the m1=m2{{m}_{1}}={{m}_{2}} (Two lines of equations are parallel) then you can find the value of p. So, the above solution is preferred for such types of problems.