Question

The line $p\left( {{p}^{2}}+1 \right)x-y+q=0$ and the line ${{\left( {{p}^{2}}+1 \right)}^{2}}x+\left( {{p}^{2}}+1 \right)y+2q=0$ are perpendicular to common line for
$\begin{aligned} & \text{a) exactly one value of p} \\\ & \text{b) exactly two values of p} \\\ & \text{c) more than two values of p} \\\ & \text{d) no values of p} \\\ \end{aligned}$

Answer 1

Now we know that if both lines are perpendicular to a common line then they are parallel to each other. Hence we have the two given lines are parallel to each other. Now we know that the lines are parallel then the slope of the lines are equal. Hence we now know that the slopes of lines are equal. Now we will write both equations in the form $y=mx+c$ . We know for a line in the form of $y=mx+c$ the slope is m. Hence we can easily find the slope of both lines and equate it to find the condition of p.

Complete step-by-step answer :
Now we are given that line $p\left( {{p}^{2}}+1 \right)x-y+q=0$ and the line ${{\left( {{p}^{2}}+1 \right)}^{2}}x+\left( {{p}^{2}}+1 \right)y+2q=0$ are perpendicular to common line.
Now we know that if a line makes same angle with a common line then the lines are parallel.
Hence we have the given two line $p\left( {{p}^{2}}+1 \right)x-y+q=0$ and ${{\left( {{p}^{2}}+1 \right)}^{2}}x+\left( {{p}^{2}}+1 \right)y+2q=0$ are parallel. ………………… (1)
Now consider the line $p\left( {{p}^{2}}+1 \right)x-y+q=0$
Let us rearrange the equation of the line by taking y to RHS we get.
$\begin{aligned} & p\left( {{p}^{2}}+1 \right)x-y+q=0 \\\ & p\left( {{p}^{2}}+1 \right)x+q=y \\\ & \Rightarrow y=p\left( {{p}^{2}}+1 \right)x+q............(2) \\\ \end{aligned}$
Now we can see that the above line is in the form of $y=mx+c$ and we know that the slope of line $y=mx+c$ is given by m. hence from equation (2) we can say
Hence the slope of line $p\left( {{p}^{2}}+1 \right)x-y+q=0$ is $p\left( {{p}^{2}}+1 \right).................(3)$
Now consider the equation ${{\left( {{p}^{2}}+1 \right)}^{2}}x+\left( {{p}^{2}}+1 \right)y+2q=0$
Dividing the whole equation by ${{p}^{2}}+1$ we get
$\dfrac{{{\left( {{p}^{2}}+1 \right)}^{2}}}{\left( {{p}^{2}}+1 \right)}x+y+\dfrac{2q}{\left( {{p}^{2}}+1 \right)}=0$
Now taking all the terms of LHS to RHS except y we get.
$y=-\dfrac{{{\left( {{p}^{2}}+1 \right)}^{2}}}{\left( {{p}^{2}}+1 \right)}x-\dfrac{2q}{\left( {{p}^{2}}+1 \right)}.............(4)$
Now we can see that the above line is in the form of $y=mx+c$ and we know that the slope of line $y=mx+c$ is given by m. hence from equation (4) we can say
Hence the slope of line ${{\left( {{p}^{2}}+1 \right)}^{2}}x+\left( {{p}^{2}}+1 \right)y+2q=0$ is $-\dfrac{{{\left( {{p}^{2}}+1 \right)}^{2}}}{\left( {{p}^{2}}+1 \right)}.................(5)$
Now if the lines are parallel we know that the slopes are equal hence from equation (4) and equation (5).
$-\dfrac{{{\left( {{p}^{2}}+1 \right)}^{2}}}{\left( {{p}^{2}}+1 \right)}=p\left( {{p}^{2}}+1 \right)$

& -\left( {{p}^{2}}+1 \right)=p\left( {{p}^{2}}+1 \right) \\\ & \Rightarrow p\left( {{p}^{2}}+1 \right)+\left( {{p}^{2}}+1 \right)=0 \\\ & \Rightarrow \left( {{p}^{2}}+1 \right)\left( p+1 \right)=0 \\\ \end{aligned}$$ Now we know that ${{p}^{2}}+1$ is never zero as $\begin{aligned} & {{p}^{2}}\ge 0 \\\ & \Rightarrow {{p}^{2}}+1>0 \\\ \end{aligned}$ Hence we get $$\begin{aligned} & p+1=0 \\\ & \Rightarrow p=-1 \\\ \end{aligned}$$ **Hence the only solution of the given equation is at p = - 1.** **Note** : Note that the equation $$\left( {{p}^{2}}+1 \right)\left( p+1 \right)=0$$ has only one root at p = - 1. Do not mistaken by writing the solution of this equation as $p=+1,-1$ .