Question

The length of the portion of the tangent to the curve $x^{\frac{2}{3}}+y^{\frac{2}{3}}=4$ at any point on it, intercepted between the coordinate axes is constant .

• A. True
• B. False

Answer 1

The correct option is A True

$x^{2/3}+y^{2/3}=4$

Slope of tangent

$\frac{2}{3}{x}^{2/3^{-1}}+\frac{2}{3}{y}^{2/3^{-1}}\frac{dy}{dx}=0$

$\frac{dy}{dx}=-\frac{x^{-1/3}}{y^{-1/3}}=\begin{array}{c}\frac{-y^{1/3}}{x^{1/3}}\end{array}|\begin{array}{c}\\ (h,k)\end{array}$

$\begin{array}{c}\frac{dy}{dx}\end{array}|\begin{array}{c}\\ (h,k)\end{array}=-{\left(\frac{k}{h}\right)}^{1/3}$

Equation of tangent at $(h,k)$

$y-k=-{\left(\frac{k}{h}\right)}^{1/3}(x-h)$

At x-axis , $y=0$

$0-k=-{\left(\frac{k}{h}\right)}^{1/3}(x-h)$

$k^{2/3}{h}^{1/3}+h=x$

$B(h+h^{1/3}{k}^{2/3},0)$

At y - axis , $x=0$

$y-k=-{\left(\frac{k}{h}\right)}^{1/3}(0-h)$

$y-k=k^{1/3}.h^{2/3}$

$y=k+k^{1/3}{h}^{2/3}$

$A(0,k+k^{1/3}{h}^{2/3})$

$(AB{)}^2=(k+k^{1/3}{h}^{2/3}-0{)}^2+(0-(h+h^{1/3}{k}^{2/3}){)}^2$

$=\left(k^{1/3}\right(k^{2/3}+h^{2/3}){)}^2+(h^{1/3}(h^{2/3}+k^{2/3}){)}^2$

$=(4k^{1/3}{)}^2+(4h^{1/3}{)}^2$

$=16k^{2/3}+16h^{2/3}$

$=16(k^{2/3}+h^{2/3})$

$\Rightarrow 16\times 4=64$

$AB=\sqrt{64}=16$