Question

If the tangent to the curve ${2y}^3={ax}^2+x^3$ at the point $(a,a)$ cuts off intercepts $\alpha$ and $\beta$ on the coordinate axes such that ${\alpha }^2+{\beta }^2=61,$ then $a=$

• A. $\pm 30$
• B. $\pm 5$
• C. $\pm 6$
• D. $\pm 61$

Answer 1

The correct option is A $\pm 30$

The slope of the tangent at any point $(a,a)$ on the curve is $m_T={\frac{dy}{dx}}_{(a,a)}$

$\therefore {6y}^2\frac{dy}{dx}=2ax+{3x}^2$

$\Rightarrow \frac{dy}{dx}=\frac{2ax+{3x}^2}{{6y}^2}$

$\Rightarrow m_T={\frac{dy}{dx}}_{(a,a)}=\frac{5}{6}$

The equation of the tangent at $(a,a)$ is $y-a=\frac{5}{6}(x-a)\Rightarrow 5x-6y+a=0$
This cuts off intercepts of lengths $-\frac{a}{5}$ and

$\frac{a}{6}$ with $x$ and $y$-axis respectively.

So, $\alpha =-\frac{a}{5}$ and $\beta =\frac{a}{6}.$
Since, ${\alpha }^2+{\beta }^2=61$ {given}

$\therefore \frac{a^2}{25}+\frac{a^2}{36}=61\Rightarrow a^2=25\times 36\therefore a=\pm 30.$