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:
\[
y=f(x)=ax^2+bx+c
\]
We are given that:
1. The curve passes through \((-1,0)\)
2. The curve touches the line \(y=x\) at \((1,1)\)
3. We need the x-intercept of the normal at the point \((\alpha,\alpha+1)\) in the first quadrant
Step 1: Use the condition that the curve passes through \((-1,0)\).
Substituting \((x,y)=(-1,0)\) in
\[
y=ax^2+bx+c
\]
we get
\[
0=a(-1)^2+b(-1)+c
\]
\[
0=a-b+c
\]
\[
a-b+c=0 \quad \cdots (1)
\]
Step 2: Use the condition that the curve touches the line \(y=x\) at \((1,1)\).
Since the curve touches the line \(y=x\) at \((1,1)\), two things must happen:
- The point \((1,1)\) lies on the curve
- The slope of the curve at \(x=1\) must be equal to the slope of the line \(y=x\), which is \(1\)
First, substitute \((1,1)\) into the quadratic:
\[
1=a(1)^2+b(1)+c
\]
\[
a+b+c=1 \quad \cdots (2)
\]
Now differentiate:
\[
y=ax^2+bx+c
\]
\[
\frac{dy}{dx}=2ax+b
\]
At \(x=1\), slope must be \(1\):
\[
2a+b=1 \quad \cdots (3)
\]
Step 3: Solve for \(a\), \(b\), and \(c\).
From equation (2) and equation (1):
\[
(a+b+c)-(a-b+c)=1-0
\]
\[
2b=1
\]
\[
b=\frac{1}{2}
\]
Substitute into equation (3):
\[
2a+\frac{1}{2}=1
\]
\[
2a=\frac{1}{2}
\]
\[
a=\frac{1}{4}
\]
Now substitute \(a=\frac14\) and \(b=\frac12\) into equation (2):
\[
\frac14+\frac12+c=1
\]
\[
\frac34+c=1
\]
\[
c=\frac14
\]
Therefore,
\[
f(x)=\frac14x^2+\frac12x+\frac14
\]
This can be written as:
\[
f(x)=\frac{(x+1)^2}{4}
\]
Step 4: Find the point \((\alpha,\alpha+1)\) on the curve in the first quadrant.
Since \((\alpha,\alpha+1)\) lies on the curve,
\[
\alpha+1=\frac{(\alpha+1)^2}{4}
\]
Multiply both sides by \(4\):
\[
4(\alpha+1)=(\alpha+1)^2
\]
\[
(\alpha+1)^2-4(\alpha+1)=0
\]
\[
(\alpha+1)\big((\alpha+1)-4\big)=0
\]
\[
(\alpha+1)(\alpha-3)=0
\]
So,
\[
\alpha=-1 \quad \text{or} \quad \alpha=3
\]
Since the point is in the first quadrant, we take
\[
\alpha=3
\]
Hence the point is
\[
(3,4)
\]
Step 5: Find the slope of the tangent and the normal at \((3,4)\).
We have
\[
f'(x)=2ax+b=\frac{x+1}{2}
\]
At \(x=3\),
\[
f'(3)=\frac{3+1}{2}=2
\]
So the slope of the tangent is
\[
2
\]
Therefore, the slope of the normal is
\[
-\frac{1}{2}
\]
Step 6: Write the equation of the normal.
The normal passes through \((3,4)\) and has slope \(-\frac12\).
Using point-slope form:
\[
y-4=-\frac12(x-3)
\]
Step 7: Find the x-intercept of the normal.
The x-intercept occurs when
\[
y=0
\]
So,
\[
0-4=-\frac12(x-3)
\]
\[
-4=-\frac12(x-3)
\]
Multiply both sides by \(-2\):
\[
8=x-3
\]
\[
x=11
\]
