Let the quadratic curve passing through the point $(-1,0)$ and touching the line $y=x$ at $(1,1)$ be $y=f(x)$. Then the $x$-intercept of the normal to the curve at the point $(\alpha, \alpha+1)$ in the first quadrant is __________.

Let the quadratic curve be $f(x)=a x^2+b x+c$ <br/><br/>The curve passes through $(-1,0)$ <br/><br/>$0=a-b+c \Rightarrow a+c=b$ ..........(i) <br/><br/>The curve also passes through $(1,1)$ <br/><br/>$$ \begin{gathered} a+b+c=1 .........(ii)\\\\ 2 b=1 \Rightarrow b=\frac{1}{2} \end{gathered} $$ <br/><br/>$f^{\prime}(x)=2 a x+b$ <br/><br/>Slope tangent of curve $=f^{\prime}(x)$ at $(1,1)=2 a+b$ <br/><br/>Slope of line $y=x$ is 1 <br/><br/>$$ \begin{aligned} &\therefore 2 a+b =1 \\\\ &2 a+\frac{1}{2} =1 \Rightarrow a=\frac{1}{4} \\\\ &c =1-a-b \Rightarrow c=\frac{1}{4} \end{aligned} $$ <br/><br/>$$ \begin{aligned} & f(x)=\frac{1}{4} x^2+\frac{1}{2} x+\frac{1}{4} = \frac{(x+1)^2}{4} \\\\ & f^{\prime}(x)=\frac{2(x+1)}{4}=\frac{x+1}{2} \end{aligned} $$ <br/><br/>$$ \begin{aligned} & f(x) \text { passes through }(\alpha, \alpha+1) \\\\ & \begin{array}{l} \therefore \alpha+1=\frac{(\alpha+1)^2}{4} \\\\ \Rightarrow \alpha+1=4 \text { or } \alpha=3 \end{array} \end{aligned} $$ <br/><br/>Equation of normal at $(3,4)$ is <br/><br/>$y-4=-\frac{1}{2}(x-3)$ <br/><br/>For $x$, intercept, $y=0$ <br/><br/>$x-3=8 \text { or } x=11$ <br/><br/>Hence, the required $x$ intercept is 11 .