Let P(x, y) be a variable point on the parabo | vedclass

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(C) Let $P = (x, y)$ be a point on $y = 4x^2 + 1$. The line $PQ$ is perpendicular to $y = x$,so its slope is $-1$. The equation of line $PQ$ is $Y - y

Vedclass · Accepted Answer

$2(3x - y)^2 + (x - 3y) + 2 = 0$

Vedclass · Answer

(C) Let $P = (x, y)$ be a point on $y = 4x^2 + 1$. The line $PQ$ is perpendicular to $y = x$,so its slope is $-1$. The equation of line $PQ$ is $Y - y = -1(X - x)$,or $X + Y = x + y$. Since $Q(c, c)$ lies on $PQ$ and $y = x$,we have $c + c = x + y$,so $c = \frac{x + y}{2}$. $Q = (\frac{x + y}{2}, \frac{x + y}{2})$. $R(h, k)$ is the midpoint of $PQ$,so $h = \frac{x + c}{2} = \frac{x + \frac{x + y}{2}}{2} = \frac{3x + y}{4}$ and $k = \frac{y + c}{2} = \frac{y + \frac{x + y}{2}}{2} = \frac{x + 3y}{4}$. Solving for $x$ and $y$: $3h + k = \frac{9x + 3y + x + 3y}{4} = \frac{10x + 6y}{4} = \frac{5x + 3y}{2}$ and $h + 3k = \frac{3x + y + 3x + 9y}{4} = \frac{6x + 10y}{4} = \frac{3x + 5y}{2}$. Alternatively,$3h - k = \frac{9x + 3y - x - 3y}{4} = 2x \implies x = \frac{3h - k}{2}$. $3k - h = \frac{3x + 9y - 3x - y}{4} = 2y \implies y = \frac{3k - h}{2}$. Substitute into $y = 4x^2 + 1$: $\frac{3k - h}{2} = 4(\frac{3h - k}{2})^2 + 1$. $\frac{3k - h}{2} = (3h - k)^2 + 1$. $3k - h = 2(3h - k)^2 + 2$. $2(3h - k)^2 + (h - 3k) + 2 = 0$. Replacing $(h, k)$ with $(x, y)$,the locus is $2(3x - y)^2 + (x - 3y) + 2 = 0$.

Let $P(x, y)$ be a variable point on the parabola $y = 4x^2 + 1$. Let $Q(c, c)$ be the foot of the perpendicular drawn from $P$ to the line $y = x$. If $R(h, k)$ is the mid-point of $PQ$,then the locus of $R$ is:

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