Question

If the line $ax+y=c$, touches both the curves ${{x}^{2}}+{{y}^{2}}=1$ and \[{{y}^{2}}=4\sqrt{2}x\] , then $|c|$ is equal to:
A. $\dfrac{1}{2}$
B. $2$
C. $\sqrt{2}$
D. $\dfrac{1}{\sqrt{2}}$

Answer 1

Hint: We will start by finding the equation of tangent to both the curves that is the circle and the parabola after that we will compare the given equation of line to the equation of tangent in order to find the value of unknown.

Complete step-by-step answer:
We know that the equation of the tangent of slope $m$ to the parabola ${{y}^{2}}=4bx$ is $y=mx+\dfrac{b}{m}...........\text{ Equation 1}\text{.}$
 We have the parabola given in the question as: \[{{y}^{2}}=4\sqrt{2x}\] , comparing it with the standard equation of parabola that is ${{y}^{2}}=4bx$, we will get $b=\sqrt{2}.$
Now it is given in the question that the line $ax+y=c$ touches the given parabola and therefore is a tangent to the parabola: le’s rewrite the given equation of line as:
$y=-ax+c$ , on comparing it with the tangent equation in equation 1 : $y=mx+\dfrac{b}{m}$
We get the slope of the line $m=-a$ and $c=\left( \dfrac{b}{m} \right)$
Placing the value of $m=-a$ and $b=\sqrt{2}$ , we get $c=\left( \dfrac{\sqrt{2}}{-a} \right)$
Therefore the tangent, $y=-ax+c$ becomes $y=-ax-\dfrac{\sqrt{2}}{a}\text{ }........\text{Equation 2}\text{.}$
Now, it is given that the given line also touches the given circle: \[{{x}^{2}}+{{y}^{2}}=1\] .
We already know that a line $px+qy+s=0$ touches the circle ${{x}^{2}}+{{y}^{2}}={{r}^{2}}$ , if \[\dfrac{|s|}{\left( \sqrt{{{p}^{2}}+{{q}^{2}}} \right)}=r\text{ }..........\text{Equation 3}\text{.}\]
We have the equation of line as $y=-ax-\dfrac{\sqrt{2}}{a}\text{ }$ from equation 2, on comparing it with the standard equation of line $px+qy+s=0$ , we have $p=-a,q=1,s=\dfrac{-\sqrt{2}}{a},r=1$ , We got the value of r=1 from the given circle : \[{{x}^{2}}+{{y}^{2}}=1\] .
Putting these values in equation 3 , we will get :
\[\begin{align}
  & \Rightarrow \dfrac{|s|}{\left( \sqrt{{{p}^{2}}+{{q}^{2}}} \right)}=r \\
 & \Rightarrow \dfrac{\left| -\dfrac{\sqrt{2}}{a} \right|}{\left( \sqrt{{{\left( -a \right)}^{2}}+{{\left( 1 \right)}^{2}}} \right)}=1\Rightarrow \dfrac{\left| -\dfrac{\sqrt{2}}{a} \right|}{\left( \sqrt{1+{{a}^{2}}} \right)}=1 \\
 & \Rightarrow \sqrt{1+{{a}^{2}}}=\left| -\dfrac{\sqrt{2}}{a} \right| \\
 & \\
\end{align}\]
We will now square both the sides and get:
\[\begin{align}
  & \Rightarrow 1+{{a}^{2}}=\dfrac{2}{{{a}^{2}}} \\
 & \Rightarrow {{a}^{2}}\left( 1+{{a}^{2}} \right)=2 \\
 & \Rightarrow {{a}^{4}}+{{a}^{2}}-2=0 \\
\end{align}\]
We will now convert it into the quadratic form let us first substitute ${{a}^{2}}$ as $u$, therefore \[{{a}^{4}}+{{a}^{2}}-2=0\] will become : \[{{u}^{2}}+u-2=0\]. We will start by factorizing the equation:
\[\begin{align}
  & \Rightarrow {{u}^{2}}+u-2=0 \\
 & \Rightarrow {{u}^{2}}+2u-u-2=0 \\
 & \Rightarrow u\left( u+2 \right)-1\left( u+2 \right)=0 \\
\end{align}\]
We will be taking $\left( u+2 \right)$ out as it is common:
\[\begin{align}
  & \Rightarrow u\left( u+2 \right)-1\left( u+2 \right)=0 \\
 & \Rightarrow \left( u+2 \right)\left( u-1 \right)=0 \\
 & \Rightarrow u=-2,1 \\
\end{align}\]
Now re-substituting ${{a}^{2}}$ as $u$ , we now have the values of ${{a}^{2}}=-2,1$ .
Here, -2 is rejected as the square does not have a negative value.
Therefore, ${{a}^{2}}=1\Rightarrow a=\pm 1$
Now we found out that: $c=\left( \dfrac{\sqrt{2}}{-a} \right)$ , we will now put the value of a into this equation:
$\therefore \left| c \right|=\left( \dfrac{\sqrt{2}}{\left| a \right|} \right)\Rightarrow \left( \dfrac{\sqrt{2}}{\left| \pm 1 \right|} \right)=\sqrt{2}$
Hence, the correct option is C.

Note: Remember to use the substitution method for solving the Quartic equation (equation with the degree 4), it is not compulsory but the calculation can get really messy with a higher degree so it is better to convert it into a quadratic equation. Also negative value for ${{a}^{2}}$ is rejected as it only contains positive value.