Q61 / 180 · KCET 2013 · Mathematics
If $\sin ^{-1} a$ is the acute angle between the curves $x^{2}+y^{2}=4 x$ and $x^{2}+y^{2}=8$ at $(2,2)$, then $a$ is equal to
- A1
- B0
- Correct answerC$\frac{1}{\sqrt{2}}$
- D$\frac{\sqrt{3}}{2}$
Solution
Given curves are,
$x^{2}+y^{2}=4 x...(i)$
On differentiating w.r.t. $x$, we get
$2 x+2 y \frac{d y}{d x}=4$
$\Rightarrow \quad \frac{d y}{d x}=\frac{2-x}{y}$
At $(2,2), \quad\left(\frac{d y}{d x}\right)_{(2,2)}=\frac{2-0}{2}=0=m_{1}$
and $\quad x^{2}+y^{2}=8$
On differentiating w.r.t. $x$, we get
$\begin{aligned}
2 x+2 y \frac{d y}{d x} &=0 \\
\Rightarrow \quad \frac{d y}{d x} &=\frac{-x}{y}
\end{aligned}$
At $(2,2)$
$\left(\frac{d y}{d x}\right)_{(2,2)}=\frac{-2}{2}=-1=m_{2}$
Let $\theta$ be an acute angle between the given curves, then, we get
$\begin{array}{cc}
& \tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} \cdot m_{2}}\right|=\left|\frac{0+1}{1+0}\right|=1 \\
\Rightarrow & \tan \theta=\tan 45^{\circ} \\
\Rightarrow & \theta=45^{\circ}
\end{array}$
$\Rightarrow \quad \sin ^{-1}(a)=45^{\circ} \quad$ (given $\left.\theta=\sin ^{-1} a\right)$
$\Rightarrow \quad a=\sin 45^{\circ}$
$\therefore \quad a=\frac{1}{\sqrt{2}}$
Hence, the correct answer is option: 'C'.