Question

Find the equation of the straight lines touching both ${{x}^{2}}+{{y}^{2}}={{a}^{2}}$ and ${{y}^{2}}=4ax$ .

Answer 1

Hint: Find the centre, radius and slope of the circle and substitute in the equation of tangent with slope.

A tangent line is a line which locally touches a curve at one and only one point
We know that the equation of tangent$\Rightarrow y=mx+c..................\left( 1 \right)$
Which is also the slope intercept formula for a line $\Rightarrow y=mx+c$
Where $m$ is the slope of the line
$c$ is the y-intercept of the line
We have been given the general form of a circle
${{x}^{2}}+{{y}^{2}}={{a}^{2}}........................\left( 2 \right)$
Here the centre of the circle is $\left( 0,0 \right)$ and
Radius $=\sqrt{{{a}^{2}}}=a$
We know the equation of parabola $\Rightarrow {{y}^{2}}=4ax....................\left( 3 \right)$
The $y-$ intercept, $c=\dfrac{a}{m}$
Where $a$ is the radius of the circle and $m$ , the slope
$\therefore$ Equation $\left( 1 \right)$ can be written as
$y=mx+c$
$\Rightarrow \left( y=mx+\dfrac{a}{m} \right).................\left( 4 \right)$
The equation of tangent to circle with slope $m$ is given by the formula
$y=mx\pm a\sqrt{1+{{m}^{2}}}.......................\left( 5 \right)$
Now comparing both equation $\left( 4 \right)\And \left( 5 \right)$ and cancelling out like terms
$\begin{aligned} & mx+\dfrac{a}{m}=mx\pm a\sqrt{1+{{m}^{2}}} \\\ & \dfrac{a}{m}=\pm a\sqrt{1+{{m}^{2}}}\Rightarrow \dfrac{1}{m}=\pm \sqrt{1+{{m}^{2}}} \\\ \end{aligned}$
Now squaring on both sides, we get
$\begin{aligned} & {{\left( \dfrac{1}{m} \right)}^{2}}={{\left( \pm \sqrt{1+{{m}^{2}}} \right)}^{2}} \\\ & \dfrac{1}{{{m}^{2}}}=1+{{m}^{2}} \\\ & \\\ \end{aligned}$
Cross multiplying the above we get
$\begin{aligned} & {{m}^{2}}\left( 1+{{m}^{2}} \right)=1 \\\ & \Rightarrow {{m}^{4}}+{{m}^{2}}-1=0 \\\ \end{aligned}$
$\therefore$ We get $\left( {{m}^{2}}+1 \right)\left( {{m}^{2}}-1 \right)=0$
We can remove the term $\left( {{m}^{2}}+1 \right)$
$\therefore \left( {{m}^{2}}-1 \right)=0\Rightarrow m=\pm 1$
Now substitute the value of $m$ in equation $\left( 4 \right)$
$\begin{aligned} & y=mx+\dfrac{a}{m} \\\ & y=\pm x+a \\\ \end{aligned}$
Substitute the value of $m$ in equation$\left( 5 \right)$

& y=mx\pm a\sqrt{1+{{m}^{2}}} \\\ & y=\pm x\pm \sqrt{2a} \\\ \end{aligned}$$ i.e. we got the required equation tangent. Note: It’s important to remember the $2$ equation of tangent with which this question was solved. $\begin{aligned} & y=mx+c\text{ or }y=mx+\dfrac{a}{m} \\\ & y=mx\pm a\sqrt{1+{{m}^{2}}} \\\ \end{aligned}$ Misplacing this equation will generate wrong answers and the desired equation of tangent won’t happen.