Question

If the tangent to the curve $2y^3=a{x}^2+x^3$ at the point $(a,a)$ cuts off intercepts $\alpha$ and $\beta$ on the co-ordinate axes, where ${\alpha }^2+{\beta }^2=61$, then the possible value(s) of $a$ is/are

• A. $-30$
• B. $10$
• C. $20$
• D. $30$

Answer 1

Answer: (A), (D)

$30$

${\left(\frac{dy}{dx}\right)}_{(a,a)}=\frac{5a^2}{6a^2}=\frac{5}{6}$

$\Rightarrow$ Equation of tangent is $-5x+6y=a$
or $\frac{x}{(-a/5)}+\frac{y}{\left(a/6\right)}=1$
$\therefore \alpha =-\frac{a}{5},\beta =\frac{a}{6}$
$\Rightarrow {\left(\frac{-a}{5}\right)}^2+{\left(\frac{a}{6}\right)}^2=61$ (given)
$\Rightarrow a=\pm 30$