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 The curve passes through (-1,0) 0=a-b+c => a+c=b ..........(i) The curve also passes through (1,1) a+b+c=1 .........(ii) \ 2 b=1 => b=(1 / 2) f^prime(x)=2 a x+b Slope tangent of curve =f^prime(x) at (1,1)=2 a+b Slope of line y=x is 1 Therefore 2 a+b =1 \ 2 a+(1 / 2) =1 => a=(1 / 4) \ c =1-a-b => c=(1 / 4) f(x)=(1 / 4) x^2+(1 / 2) x+(1 / 4) = ((x+1)^2 / 4) \ f^prime(x)=(2(x+1) / 4)=(x+1 / 2) f(x) passes through (alpha, alpha+1) \ \l Therefore alpha+1=((alpha+1)^2 / 4) \ => alpha+1=4 or alpha=3 Equation of normal at (3,4) is y-4=-(1 / 2)(x-3) For x, intercept, y=0 x-3=8 or x=11 Hence, the required x intercept is 11 .

Let the quadratic curve passing through the point 10 an... | null (Mathematics)

Let the quadratic curve be

f(x)=a x^2+b x+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)\\\\ 2 b=1 \Rightarrow b=\frac{1}{2} \end{gathered}

f^{\prime}(x)=2 a x+b

Slope tangent of curve

=f^{\prime}(x)

(1,1)=2 a+b

Slope of line

y=x

is 1

\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}

\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}

\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}

Equation of normal at

(3,4)

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

For

x

, intercept,

y=0

x-3=8 \text { or } x=11

Hence, the required

x

intercept is 11 .

Explanation

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