Question
If the tangent to the curve $2 y^{3}=a x^{2}+x^{3}$ 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) $\pm 61$
Step 1
This is done by differentiating the equation with respect to $x$ to get $dy/dx$. \[ \frac{dy}{dx} = \frac{2ax + 3x^{2}}{6y^{2}} \] Show more…
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