If the sum of the squares of the intercepts on the axes cut off by the tangent to the cuve $x^{1 / 3} + y^{1 / 3} = a^{1 / 3} (a > 0)$ at $(\frac{a}{8}, \frac{a}{8})$ is $2$ , then $a$ has the value

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

8

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $4$
$x^{1 / 3} + y^{1 / 3} = a^{1 / 3} (a > 0)$
$y^{'} (x) = - \frac{y^{2 / 3}}{x^{2 / 3}}$
${y^{'} |}_{(x, y) = (\frac{a}{8}, \frac{a}{8})}^{'} = - 1$

Equation of tangent
$x + y = c$
$x + y = \frac{a}{4}; ∵ (\frac{a}{8}, \frac{a}{8})$ satisfies it

$x -$ intercept $= a \frac{a}{4} = y -$ intercept
Given, $\frac{a^{2}}{16} + \frac{a^{2}}{16} = 2$
$a = \pm 4$
$a = 4; ∵ a > 0$

Suggest Corrections

Similar questions

The sum of squares of the intercepts on the coordinate axes made by the tangent to $x^{1 / 3} + y^{1 / 3} = a^{1 / 3} a t (\frac{a}{8}, \frac{a}{8})$ is 2, and a > 0 then a = ____

If the sum of squares of the intercept on the axes cut off by the tangent on the curve $x^{\frac{1}{3}} + y^{\frac{1}{3}} = a^{\frac{1}{3}}, a > 0$ at $(\frac{a}{8}, \frac{a}{8})$ is $2,$ then value of $a$ is

For what values of a a and b the intercepts cut off on the coordinate axes by the line ax + by + 8 = 0 are equal in length but opposite in signs to those cut off by the line 2x - 3y + 6 = 0 on the axes.

Q. If the tangent to the curve