Ok here i have a calculus problem to solve to find the area of the curve:
Calculate the surface area bounded by the curves $ 3x^2-1 $ and its tangents that pass through point $(0,1/4)$.
My question is how to find the equation of tangents
2026-03-31 02:20:54.1774923654
Calculate the surface area bounded by the curves $3x^2−1$ and its tangents that pass through point (0,1/4).
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There are no tangents to the curve $y=3x^2-1$ passing through the point $A=\left(0,\frac{1}{4}\right)$
Any line passing through $A$ has equation $y=mx+1/4$ intersects the curve in two distinct points
Substitute in the curve equation $3x^2-1=mx+1/4$ and solve
$$x = \frac{m\pm\sqrt{m^2+15}}{6}$$ As $m^2+15>0$ for any $m$ the equation has always two real solutions.