Question

Let $P$ be any point on the curve $x^{2 / 3}+y^{2 / 3}=a^{2 / 3}$. Then, the length of the segment of the tangent between the coordinate axes of length

• A. 3a
• B. 4a
• C. 5a
• D. a

Answer 1

Answer: (D)

Let the coordinates of the point $P$ be $\left(x_{1}, y_{1}\right)$.
This point lies on the curve
$x^{2 / 3}+y^{2 / 3} =a^{2 / 3}$ ...(i)
$\therefore x_{1}^{2 / 3}+y_{1}^{2 / 3} =a^{2 / 3}$ ...(ii)
On differentiating Eq. (i) w.r.t. $x$, we get
$\frac{2}{3} x^{-1 / 3}+\frac{2}{3} y^{-1 / 3} \frac{d y}{d x}=0$
$\Rightarrow \frac{d y}{d x}=-\frac{y^{1 / 3}}{x^{1 / 3}}$
$\Rightarrow \left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)} =-\left(\frac{y_{1}}{x_{1}}\right)^{1 / 3}$
The equation of the tangent at $\left(x_{1}, y_{1}\right)$ to the given curve is
$y-y_{1}=-\frac{y_{1}^{1 / 3}}{x_{1}^{1 / 3}}\left(x-x_{1}\right)$
$\Rightarrow x x_{1}^{-1 / 3}+y y_{1}^{-1 / 3}=x_{1}^{2 / 3}+y_{1}^{2 / 3}$
$\Rightarrow x x_{1}^{-1 / 3}+y y_{1}^{-1 / 3}=a^{2 / 3}$ [using Eq. (ii)]
This tangent meets the coordinate axes at
$A\left(a^{2 / 3} x_{1}^{1 / 3}, 0\right)$ and $B\left(0, a^{2 / 3} y_{1}^{1 / 3}\right)$
$\therefore A B =\sqrt{\left(0-a^{2 / 3} x_{1}^{1 / 3}\right)^{2}+\left(a^{2 / 3} y_{1}^{1 / 3}-0\right)^{2}}$
$=\sqrt{a^{4 / 3}\left(x_{1}^{2 / 3}+y_{1}^{2 / 3}\right)}$ [from Eq. (i)]
$=\sqrt{a^{4 / 3} \cdot a^{2 / 3}}$
$=\sqrt{a^{2}}=a$