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

The correct option is D $a$

Let the coordinates of the point $P$ be $(x_1,y_1)$.
This point lies on the curve
$x^{2/3}+y^{2/3}=a^{2/3}$ .... (i)
and $x_1^{2/3}+y_1^{2/3}=a^{2/3}$ .... (ii)
On differentiating Eq. (i) w.r.t. $x$, we get
$\frac{2}{3}{x}^{-1/3}+\frac{2}{3}{y}^{-1/3}\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=\frac{y^{1/3}}{x^{1/3}}$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{(x_1,y_1)}=-{\left(\frac{y_1}{x_1}\right)}^{1/3}$
The equation of the tangent at $(x_1,y_1)$ to the given curve is
$y-y_1=-\frac{y_1^{1/3}}{x_1^{1/3}}(x-x_1)$
$\Rightarrow x{x}_1^{-1/3}+y{y}_1^{-1/3}=x_1^{2/3}+y_1^{2/3}$
$\Rightarrow x{x}_1^{-1/3}+y{y}_1^{-1/3}=a^{2/3}$ ....[using Eq. (ii)]
This tangent meets the coordinate axes at $A(a^{2/3}{x}_1^{1/3},0)$ and $B(0,a^{2/3}{y}_1^{1/3})$.
$AB=\sqrt{(0-a^{2/3}{x}_1^{1/3}{)}^2+(a^{2/3}{y}_1^{1/3}-0{)}^2}$
$=\sqrt{a^{4/3}(x_1^{2/3}+y_1^{2/3})}$ ....[from Eq. (ii)]
$=\sqrt{a^{4/3}\cdot a^{2/3}}$
$=\sqrt{a^2}=a$