Question

What is the normal to the y^2=4c(x-b)

Answer 1

tx + y = (b+2c)t + ct^3

Answer 2

To find the normal to the curve $y^2 = 4c(x-b)$, we first determine the general point on the parabola and then find the slope of the tangent and subsequently the normal.

The given equation is $y^2 = 4c(x-b)$. This is a parabola with its vertex at $(b,0)$ and its axis along the x-axis.

1. Parametric form of the parabola:

Let $X = x-b$. The equation becomes $y^2 = 4cX$.

The standard parametric form for a parabola $y^2 = 4aX$ is $(X, y) = (at^2, 2at)$.

In our case, $a=c$, so the parametric coordinates for $(X, y)$ are $(ct^2, 2ct)$.

Substituting $X = x-b$, we get:

$x-b = ct^2 \implies x = b + ct^2$

$y = 2ct$

So, any point on the parabola can be represented as $(b+ct^2, 2ct)$.

2. Slope of the tangent ($\frac{dy}{dx}$):

We find $\frac{dy}{dx}$ using parametric differentiation:

$\frac{dx}{dt} = \frac{d}{dt}(b+ct^2) = 2ct$

$\frac{dy}{dt} = \frac{d}{dt}(2ct) = 2c$

Now, $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2c}{2ct} = \frac{1}{t}$ (for $t \neq 0$).

3. Slope of the normal:

The slope of the normal ($m_N$) is the negative reciprocal of the slope of the tangent ($m_T$):

$m_N = -\frac{1}{m_T} = -\frac{1}{1/t} = -t$.

4. Equation of the normal:

Using the point-slope form of a line, $Y - Y_1 = m(X - X_1)$, where $(X_1, Y_1) = (b+ct^2, 2ct)$ and $m = -t$:

$y - 2ct = -t(x - (b+ct^2))$

5. Simplify the equation:

$y - 2ct = -tx + bt + ct^3$

Rearranging the terms to a standard form:

$tx + y = bt + 2ct + ct^3$

$tx + y = (b+2c)t + ct^3$

This is the general equation of the normal to the parabola $y^2 = 4c(x-b)$ in parametric form, where 't' is the parameter.