Question

If $q$is the acute angle of intersection at a real point of intersection of the circle ${{x}^{2}}+{{y}^{2}}=5$ and the parabola ${{y}^{2}}=4x$ then $\tan q$ is equal to
A. $1$
B. $\sqrt{3}$
C. $3$
D. $\dfrac{1}{\sqrt{3}}$

Answer 1

Hint: To solve this question we will solve both the given equations with the help of substitution method and determine the value of $x$ and $y$and make the points of intersection. We will then differentiate first equation with respect to $x$ at the point of intersection and determine the value of ${{m}_{1}}$. We will differentiate the second equation at the point of intersection and determine the value of ${{m}_{2}}$. We will then substitute the value of ${{m}_{1}}$ and ${{m}_{2}}$in the formula of $\tan A$ and determine its value.



Formula Used:$\tan A=\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|$
$m=\dfrac{dy}{dx}$





Complete step by step solution:We are given that $q$is the acute angle of intersection at a real point of intersection of the circle ${{x}^{2}}+{{y}^{2}}=5$ and the parabola ${{y}^{2}}=4x$ and we have to find the value of $\tan q$.
We will substitute the equation ${{y}^{2}}=4x$ in equation ${{x}^{2}}+{{y}^{2}}=5$ and derive the value of $x$ and $y$.
$\begin{align}
  & {{x}^{2}}+{{y}^{2}}=5 \\
 & {{x}^{2}}+4x=5 \\
 & {{x}^{2}}+4x-5=0
\end{align}$
Factorizing the quadratic equation.
$\begin{align}
  & {{x}^{2}}+4x-5=0 \\
 & x(x+5)-(x+5)=0 \\
 & (x+5)(x-1)=0 \\
 & x=1,-5
\end{align}$
We will now find the value of $y$by substituting both the values of $x$in equation ${{y}^{2}}=4x$.
$\begin{align}
  & {{y}^{2}}=4(1) \\
 & {{y}^{2}}=4 \\
 & y=\pm 2
\end{align}$ Or $\begin{align}
  & {{y}^{2}}=4(-5) \\
 & {{y}^{2}}=-20
\end{align}$
We will consider value of $y$ as $y=\pm 2$ . So the points of intersection will be $(1,2)$ and $(1,-2)$.
Now,
We will differentiate the equation ${{x}^{2}}+{{y}^{2}}=5$ with respect to $x$ to find the value of first slope ${{m}_{1}}$ at point $(1,2)$.
$\begin{align}
  & {{x}^{2}}+{{y}^{2}}=5 \\
 & 2x+2y\dfrac{dy}{dx}=0 \\
 & 2y\dfrac{dy}{dx}=-2x \\
 & \dfrac{dy}{dx}=-\dfrac{x}{y}
\end{align}$
Hence the value of ${{m}_{1}}$ will be,
$\begin{align}
  & {{m}_{1}}={{\left( \dfrac{dy}{dx} \right)}_{\left( 1,2 \right)}} \\
 & ={{\left( \dfrac{-x}{y} \right)}_{\left( 1,2 \right)}} \\
 & =-\dfrac{1}{2}
\end{align}$
We will now differentiate the equation ${{y}^{2}}=4x$ with respect to $x$ to find the value of second slope ${{m}_{2}}$ at $(1,2)$.
$\begin{align}
  & {{y}^{2}}=4x \\
 & 2y\dfrac{dy}{dx}=4 \\
 & \dfrac{dy}{dx}=\dfrac{4}{2y} \\
 & \dfrac{dy}{dx}=\dfrac{2}{y}
\end{align}$
Hence the value of ${{m}_{2}}$ will be,
$\begin{align}
  & {{m}_{2}}={{\left( \dfrac{dy}{dx} \right)}_{\left( 1,2 \right)}} \\
 & ={{\left( \dfrac{2}{y} \right)}_{\left( 1,2 \right)}} \\
 & =\dfrac{2}{2} \\
 & =1
\end{align}$
We will now substitute the value of ${{m}_{2}}$ and ${{m}_{1}}$ in the formula of $\tan A$ and determine the value of $\tan q$.
$\begin{align}
  & \tan q=\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right| \\
 & =\left| \dfrac{-\dfrac{1}{2}-1}{1+\left( -\dfrac{1}{2} \right)1} \right| \\
 & =\left| \dfrac{-\dfrac{3}{2}}{\dfrac{1}{2}} \right| \\
 & =3
\end{align}$
The value of $\tan q$ is $3$ where $q$is the acute angle of intersection at a real point of intersection of the circle ${{x}^{2}}+{{y}^{2}}=5$ and the parabola ${{y}^{2}}=4x$.



Option ‘C’ is correct



Note: Slope can be defined as the ratio of change in the values of $y$over the change in the values of $x$.