The aim of this paper is analyzing the positive solutions of the quasilinear problem - ( u ′ / 1 + ( u ′ ) 2 ) ′ = λ a ( x ) f ( u ) in ( 0 , 1 ) , u ′ ( 0 ) = 0 , u ′ ( 1 ) = 0 , -\bigl{(}u^{\prime}/\sqrt{1+(u^{\prime})^{2}}\big{)}^{\prime}=\lambda a(x)f(u)% \quad\text{in }(0,1),\qquad u^{\prime}(0)=0,\quad u^{\prime}(1)=0, where λ∈ℝ{\lambda\in\mathbb{R}} is a parameter, a∈L∞(0,1){a\in L^{\infty}(0,1)} changes sign once in (0,1){(0,1)} and satisfies ∫01a(x)𝑑x<0{\int_{0}^{1}a(x)\,dx<0}, and f∈𝒞1(ℝ){f\in\mathcal{C}^{1}(\mathbb{R})} is positive and increasing in (0,+∞){(0,+\infty)} with a potential, F(s)=∫0sf(t)𝑑t{F(s)=\int_{0}^{s}f(t)\,dt}, quadratic at zero and linear at +∞{+\infty}. The main result of this paper establishes that this problem possesses a component of positive bounded variation solutions, 𝒞λ0+{\mathscr{C}_{\lambda_{0}}^{+}}, bifurcating from (λ,0){(\lambda,0)} at some λ0>0{\lambda_{0}>0} and from (λ,∞){(\lambda,\infty)} at some λ∞>0{\lambda_{\infty}>0}. It also establishes that 𝒞λ0+{\mathscr{C}_{\lambda_{0}}^{+}} consists of regular solutions if and only if ∫ 0 z ( ∫ x z a ( t ) d t ) - 1 / 2 d x = + ∞ or ∫ z 1 ( ∫ x z a ( t ) d t ) - 1 / 2 d x = + ∞ . \int_{0}^{z}\Biggr{(}\int_{x}^{z}a(t)\,dt\Bigg{)}^{-{1/2}}dx=+\infty\quad\text% {or}\quad\int_{z}^{1}\Biggr{(}\int_{x}^{z}a(t)\,dt\Bigg{)}^{-{1/2}}dx=+\infty. Equivalently, the small positive regular solutions of 𝒞λ0+{\mathscr{C}_{\lambda_{0}}^{+}} become singular as they are sufficiently large if and only if ( ∫ x z a ( t ) d t ) - 1 / 2 ∈ L 1 ( 0 , z ) and ( ∫ x z a ( t ) d t ) - 1 / 2 ∈ L 1 ( z , 1 ) . \Biggr{(}\int_{x}^{z}a(t)\,dt\Bigg{)}^{-{1/2}}\in L^{1}(0,z)\quad\text{and}% \quad\Biggr{(}\int_{x}^{z}a(t)\,dt\Bigg{)}^{-{1/2}}\in L^{1}(z,1). This is achieved by providing a very sharp description of the asymptotic profile, as λ→λ∞{\lambda\to\lambda_{\infty}}, of the solutions. According to the mutual positions of λ0{\lambda_{0}} and λ∞{\lambda_{\infty}}, as well as the bifurcation direction, the occurrence of multiple solutions can also be detected.