Let the quadratic curve passing through the point (-1,0) and touching the line y=x at (1,1…

Question

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 \_\_\_\_\_\_\_\_...

Answer: 11

Accepted Answer

Let the quadratic curve be $f(x)=a{x}^{2}+bx+c$

The curve passes through $(-1,0)$

$0=a-b+c\Rightarrow a+c=b$ ..........(i)

The curve also passes through $(1,1)$

$\begin{gathered} a+b+c=1.........(ii) \ 2b=1\Rightarrow b=\frac{1}{2} \end{gathered}$

${f}^{'}(x)=2ax+b$

Slope tangent of curve $={f}^{'}(x)$ at $(1,1)=2a+b$

Slope of line $y=x$ is 1

$\begin{gathered} 	herefore 2a+b=1 \ 2a+\frac{1}{2}=1\Rightarrow a=\frac{1}{4} \ c=1-a-b\Rightarrow c=\frac{1}{4} \end{gathered}$

$\begin{gathered} f(x)=\frac{1}{4}{x}^{2}+\frac{1}{2}x+\frac{1}{4}=\frac{(x+1{)}^{2}}{4} \ {f}^{'}(x)=\frac{2(x+1)}{4}=\frac{x+1}{2} \end{gathered}$

$\begin{gathered} f(x)(\alpha ,\alpha +1) \ \begin{gathered} 	herefore \alpha +1=\frac{(\alpha +1{)}^{2}}{4} \ \Rightarrow \alpha +1=4\alpha =3 \end{gathered} \end{gathered}$

Equation of normal at $(3,4)$ is

$y-4=-\frac{1}{2}(x-3)$

For $x$, intercept, $y=0$

$x-3=8x=11$

Hence, the required $x$ intercept is 11 .

Solutions · 1