Question

A line L is common tangent to the circle ${{x}^{2}}+{{y}^{2}}=1$ and the parabola ${{y}^{2}}=4x$. If $\theta $ is the angle which it makes with the positive x-axis, then ${{\tan }^{2}}\theta $ is equal to
(a) \[2\sin {{18}^{\circ }}\]
(b) \[2\sin {{15}^{\circ }}\]
(c) \[\cos {{36}^{\circ }}\]
(d) \[2\cos {{36}^{\circ }}\]

Answer 1

Hint: By comparing the given equations for the parabola and the circle with their respective standard equations \[{{y}^{2}}=4ax\] and ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$ we can write $a=1$, $h=0$, $k=0$ and $r=1$. Then, these values have to be substituted into the general equations for the tangent to the parabola and the circle, which are respectively given by $y=mx+\dfrac{a}{m}$ and $y-k=m\left( x-h \right)\pm r\sqrt{{{m}^{2}}+1}$. Then on equating the constant terms in these equations, we will get the equation for the slope of the common tangent, which is equal to $\tan \theta $.

Complete step by step solution:
The equation for the parabola is given in the above question as
\[\Rightarrow {{y}^{2}}=4x\]
On comparing the above equation by the standard equation of a parabola \[{{y}^{2}}=4ax\], we get
$\Rightarrow a=1$
Now, we know that the equation for the tangent to a parabola is given as
$\Rightarrow y=mx+\dfrac{a}{m}$
Substituting $a=1$ we get
$\Rightarrow y=mx+\dfrac{1}{m}......\left( i \right)$
Now, the equation for the circle is given as
$\Rightarrow {{x}^{2}}+{{y}^{2}}=1$
On comparing the above equation with the standard equation for a circle ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$, we get
$\begin{align}
  & \Rightarrow h=0.......\left( ii \right) \\
 & \Rightarrow k=0.......\left( iii \right) \\
 & \Rightarrow r=1.......\left( iv \right) \\
\end{align}$
Now, we know that the general equation for the tangent to a circle is given as
$\Rightarrow y-k=m\left( x-h \right)\pm r\sqrt{{{m}^{2}}+1}$
Substituting the equations (ii), (iii) and (iv) in the above equation, we get
$\begin{align}
  & \Rightarrow y-0=m\left( x-0 \right)\pm 1\sqrt{{{m}^{2}}+1} \\
 & \Rightarrow y=mx\pm \sqrt{{{m}^{2}}+1}......\left( v \right) \\
\end{align}$
Since the line L is a common tangent to the given circle and the parabola, we can equate the constant terms in the equations (i) and (v) to get
$\Rightarrow \dfrac{1}{m}=\pm \sqrt{{{m}^{2}}+1}$
On squaring both the sides, we get
$\begin{align}
  & \Rightarrow {{\left( \dfrac{1}{m} \right)}^{2}}={{\left( \pm \sqrt{{{m}^{2}}+1} \right)}^{2}} \\
 & \Rightarrow \dfrac{1}{{{m}^{2}}}=\left( {{m}^{2}}+1 \right) \\
\end{align}$
Multiplying the above equation by \[{{m}^{2}}\] we get
\[\begin{align}
  & \Rightarrow \dfrac{1}{{{m}^{2}}}\times {{m}^{2}}={{m}^{2}}\left( {{m}^{2}}+1 \right) \\
 & \Rightarrow 1={{m}^{2}}\left( {{m}^{2}}+1 \right) \\
 & \Rightarrow {{m}^{2}}\left( {{m}^{2}}+1 \right)=1 \\
\end{align}\]
Using the distributive law, we can simplify the LHS of the above equation as
$\begin{align}
  & \Rightarrow {{m}^{2}}\left( {{m}^{2}} \right)+{{m}^{2}}=1 \\
 & \Rightarrow {{\left( {{m}^{2}} \right)}^{2}}+{{m}^{2}}=1 \\
\end{align}$
Subtracting $1$ from both the sides, we get
\[\begin{align}
  & \Rightarrow {{\left( {{m}^{2}} \right)}^{2}}+{{m}^{2}}-1=1-1 \\
 & \Rightarrow {{\left( {{m}^{2}} \right)}^{2}}+{{m}^{2}}-1=0 \\
\end{align}\]
On substituting ${{m}^{2}}=y$ we can write the above equation as
$\Rightarrow {{y}^{2}}+y-1=0......\left( vi \right)$
Now, we know that the quadratic formula is given by
$\Rightarrow y=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
From the equation (vi) we can substitute $a=1,b=1,c=-1$ to get
\[\begin{align}
  & \Rightarrow y=\dfrac{-1\pm \sqrt{{{\left( 1 \right)}^{2}}-4\left( 1 \right)\left( -1 \right)}}{2\left( 1 \right)} \\
 & \Rightarrow y=\dfrac{-1\pm \sqrt{1+4}}{2} \\
 & \Rightarrow y=\dfrac{-1\pm \sqrt{5}}{2} \\
 & \Rightarrow y=\dfrac{-1+\sqrt{5}}{2},y=\dfrac{-1-\sqrt{5}}{2} \\
\end{align}\]
Since $y={{m}^{2}}$, it cannot be negative. Therefore, \[y=\dfrac{-1-\sqrt{5}}{2}\] is rejected. So we have
$\Rightarrow y=\dfrac{-1+\sqrt{5}}{2}$
Now, we substitute $y={{m}^{2}}$ back into the above equation to get
\[\begin{align}
  & \Rightarrow {{m}^{2}}=\dfrac{-1+\sqrt{5}}{2} \\
 & \Rightarrow {{m}^{2}}=\dfrac{\sqrt{5}-1}{2} \\
\end{align}\]
Now, according to the question, $\theta $ is the angle which the common tangent makes with the positive x-axis. We know that the slope of a line is equal to the tangent of the angle made by the line with the positive x-axis. Therefore, we can substitute $m=\tan \theta $ in the above equation to get
$\Rightarrow {{\tan }^{2}}\theta =\dfrac{\sqrt{5}-1}{2}......\left( vii \right)$
We know that
$\Rightarrow \sin {{18}^{\circ }}=\dfrac{\sqrt{5}-1}{4}$
Multiplying by two on both the sides, we get
\[\begin{align}
  & \Rightarrow 2\sin {{18}^{\circ }}=2\left( \dfrac{\sqrt{5}-1}{4} \right) \\
 & \Rightarrow 2\sin {{18}^{\circ }}=\dfrac{\sqrt{5}-1}{2} \\
 & \Rightarrow \dfrac{\sqrt{5}-1}{2}=2\sin {{18}^{\circ }} \\
\end{align}\]
Substituting this in the equation (vii) we finally get
\[\Rightarrow {{\tan }^{2}}\theta =2\sin {{18}^{\circ }}\]
Hence, the correct answer is option (a).

Note: The value of \[\sin {{18}^{\circ }}\] is an important value and so it must be remembered. Also, for solving these types of questions, we must remember the general equations for all the conic curves. Otherwise, it will involve a lot of calculations for deriving the equation for the tangent on such curves.