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Step-by-Step Solution
Step 1: Identify the Parabola and the Line
We are given the parabola
$y = 4x^2 + 1$,
and the line
$y = x$.
Step 2: Define the Points
• Let
$P(x, y)$
be a variable point on the parabola
$y = 4x^2 + 1$.
• Let
$Q(q, q)$
be the foot of the perpendicular from
$P(x, y)$
to the line
$y = x$,
because any point on the line
$y = x$
has coordinates of the form
$(q, q)$.
• Let
$R(a, b)$
be the midpoint of the segment
$PQ$.
Step 3: Express the Midpoint Coordinates
Since
$R(a, b)$
is the midpoint of
$P(x, y)$
and
$Q(q, q)$,
by the midpoint formula we have:
$a = \frac{x + q}{2}, \quad b = \frac{y + q}{2}.$
Step 4: Use the Perpendicularity Condition
The line
$y = x$
has slope
$m_1 = 1$.
If
$PQ$
is perpendicular to
$y = x$,
its slope
$m_2$
must satisfy
$m_1 \cdot m_2 = -1$.
That is,
$\text{slope of } PQ = m_2 = \frac{b - q}{a - q}.$
Therefore,
$1 \cdot \frac{b - q}{a - q} = -1
\quad \Longrightarrow \quad \frac{b - q}{a - q} = -1.$
From this equation,
$b - q = -(a - q)
\quad \Longrightarrow \quad b - q = q - a
\quad \Longrightarrow \quad q = \frac{a + b}{2}.$
Step 5: Express x and y in Terms of a and b
Using
$q = \frac{a + b}{2},
\quad a = \frac{x + q}{2},
\quad b = \frac{y + q}{2},$
we solve for
$x$
and
$y$.
Begin with
$a = \frac{x + q}{2}$:
$a = \frac{x + \frac{a + b}{2}}{2}
\; \Longrightarrow \; 2a = x + \frac{a + b}{2}
\; \Longrightarrow \; 4a = 2x + a + b
\; \Longrightarrow \; 3a - b = 2x
\; \Longrightarrow \; x = \frac{3a - b}{2}.$
Next, use
$b = \frac{y + q}{2}$:
$b = \frac{y + \frac{a + b}{2}}{2}
\; \Longrightarrow \; 2b = y + \frac{a + b}{2}
\; \Longrightarrow \; 4b = 2y + a + b
\; \Longrightarrow \; 3b - a = 2y
\; \Longrightarrow \; y = \frac{3b - a}{2}.$
Step 6: Substitute into the Parabola Equation
Because
$P(x, y)$
lies on
$y = 4x^2 + 1$,
we substitute
$x = \frac{3a - b}{2}$
and
$y = \frac{3b - a}{2}$
into the parabola:
$\frac{3b - a}{2} = 4 \left(\frac{3a - b}{2}\right)^2 + 1.$
Simplifying,
$\frac{3b - a}{2} = 4 \cdot \frac{(3a - b)^2}{4} + 1
= (3a - b)^2 + 1.$
Multiply both sides by 2:
$3b - a = 2(3a - b)^2 + 2.$
Step 7: Obtain the Locus in Terms of (a, b)
Rewrite in standard form:
$3b - a - 2 = 2(3a - b)^2.$
Or equivalently,
$2(3a - b)^2 + (a - 3b) + 2 = 0.$
Since
$(a, b)$
was our midpoint (now interpreted as the locus variables
$(x, y)$),
replace
$(a, b)$
by
$(x, y)$
to get the locus equation:
$2(3x - y)^2 + (x - 3y) + 2 = 0.$
Final Answer
The locus of the midpoint
$R(a, b)$
is
$2(3x - y)^2 + (x - 3y) + 2 = 0.$
