Q107 / 160 · TS EAMCET 2016 · Mathematics
The length of the segment of the tangent line to the curve $x=a \cos ^3 t, y=\alpha \sin ^3 t$, at any point on the curve cut off by the coordinate axes is
- A$4 a$
- Correct answerB$a$
- C$a^2$
- D$2 a$
Solution
Given, curves are
and
$$
\begin{aligned}
& x=a \cos ^3 t \\
& y=a \sin ^3 t
\end{aligned}
$$
On differentiating Eq. (i) and Eq. (ii) w.r.t.t, we get
and
$$
\begin{aligned}
& \frac{d y}{d t}=3 a \sin ^2 t \cos t \\
& \frac{d x}{d t}=-3 a \cos ^2 t \sin t
\end{aligned}
$$
$$
\begin{aligned}
\therefore \quad \frac{d y}{d x} & =\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{3 a \sin ^2 t \cos t}{-3 a \cos ^2 t \sin t} \\
& =\frac{-\sin t}{\cos t}=m_1
\end{aligned}
$$
Equation of tangent at point $\left(\operatorname{acos}^3 t, a \sin ^3 t\right)$ is
$$
\begin{gathered}
\quad y-a \sin ^3 t=\frac{-\sin t}{\cos t}\left(x-a \cos ^3 t\right) \\
\Rightarrow y \cos t-a \cos t \sin ^3 t=-x \sin t+a \cos ^3 t \sin t \\
\Rightarrow x \sin t+y \cos t=a \cos t \sin t\left[\sin ^2 t+\cos ^2 t\right] \\
\Rightarrow x \sin t+y \cos t=a \cos t \sin t
\end{gathered}
$$
For finding the points cut off by the coordinate axes put $y=0$ in Eq. (iii), we get
$$
\begin{aligned}
& \Rightarrow \quad x \sin t=\operatorname{acos} t \sin t \\
& \Rightarrow \quad x=a \cos t \\
& \text { and put } \\
& x=0 \text { in Eq. (iii), we get } \\
& \Rightarrow \quad y \cos t=a \cos t \sin t \\
& \Rightarrow \quad y=a \sin t \\
&
\end{aligned}
$$
Hence, points are $A(a \cos t, 0)$ and $B(0, a \sin t)$,
$$
\begin{aligned}
\therefore \quad \text { Length of } & A B=\sqrt{(a \cos t)^2+(a \sin f)^2} \\
& =\sqrt{a^2\left(\cos ^2 t+\sin ^2 t\right)}=\sqrt{a^2}=a
\end{aligned}
$$
Hence, the correct answer is option: 'B'.