• # question_answer A chord of the parabola $y=-\,{{a}^{2}}{{x}^{2}}+5ax-4$ touches the curve $y=\frac{1}{1-x}$ at the point $x=2$ and is bisected by that point. If S is the sum of all possible values of a, then find 12S. A) 12 B) 15 C) 17 D) 19

 Slope of the tangent ${{\left. \frac{dy}{dx}=\frac{1}{{{(1-x)}^{2}}}\, \right|}_{x\,=\,2}}=1$ Equation of tangent is $y+1=1\,(x-2)$ i.e.        $y=x-3$ parabola $y=-{{a}^{2}}{{x}^{2}}+5ax-4$ Solving the equations of tangent and the parabola $x-3=-{{a}^{2}}{{x}^{2}}+5ax-4$ ${{a}^{2}}{{x}^{2}}+(1-5a)\,x+1=0$ Since x is real     $\therefore$      ${{(1-5a)}^{2}}-4{{a}^{2}}\ge 0$ $\Rightarrow$$a\le \frac{1}{7}$ or $a\ge \frac{1}{3}$ Sum of roots  $=\frac{5a-1}{{{a}^{2}}}=2\times 2$ $4{{a}^{2}}-5a+1=0$ $a=1,$$\frac{1}{4}$                [$a=\frac{1}{4}$is rejected] $\therefore$      $S=1$              $\therefore$$12S=12$