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
$\frac{2}{3}{x}^{-1/3}+\frac{2}{3}{y}^{-1/3}\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=-\frac{y^{1/3}}{x^{1/3}}$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}=-{\left(\frac{y_1}{x_1}\right)}^{1/3}$
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 $B\left(0,a^{2/3}{y}_1^{1/3}\right)$
$\therefore AB=\sqrt{{\left(0-a^{2/3}{x}_1^{1/3}\right)}^2+{\left(a^{2/3}{y}_1^{1/3}-0\right)}^2}$
$=\sqrt{a^{4/3}\left(x_1^{2/3}+y_1^{2/3}\right)}$ [from Eq. (i)]
$=\sqrt{a^{4/3}\cdot a^{2/3}}$
$=\sqrt{a^2}=a$