To find the equations of the tangent and normal to the parabola [tex]\( y^2 = 16ax \)[/tex] at the point where the ordinate (the [tex]\(y\)[/tex]-coordinate) is [tex]\(-4a\)[/tex], follow these steps:
### Step 1: Find the Point of Tangency
Given the ordinate [tex]\( y = -4a \)[/tex], substitute this into the parabola equation to find the corresponding [tex]\( x \)[/tex]-coordinate:
[tex]\[ y^2 = 16ax \][/tex]
[tex]\[ (-4a)^2 = 16ax \][/tex]
[tex]\[ 16a^2 = 16ax \][/tex]
[tex]\[ x = a \][/tex]
So, the point of tangency is [tex]\( (a, -4a) \)[/tex].
### Step 2: Find the Slope of the Tangent
To find the slope of the tangent to the parabola at any point [tex]\((x_1, y_1)\)[/tex], differentiate the equation [tex]\( y^2 = 16ax \)[/tex] implicitly with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{d}{dx}(y^2) = \frac{d}{dx}(16ax) \][/tex]
[tex]\[ 2y \frac{dy}{dx} = 16a \][/tex]
[tex]\[ \frac{dy}{dx} = \frac{16a}{2y} \][/tex]
[tex]\[ \frac{dy}{dx} = \frac{8a}{y} \][/tex]
At the point [tex]\( (a, -4a) \)[/tex]:
[tex]\[ \frac{dy}{dx} \bigg|_{(a, -4a)} = \frac{8a}{-4a} = -2 \][/tex]
So, the slope [tex]\( m \)[/tex] of the tangent at [tex]\( (a, -4a) \)[/tex] is [tex]\(-2\)[/tex].
### Step 3: Find the Equation of the Tangent Line
Using the point-slope form of the equation of a line [tex]\( y - y_1 = m(x - x_1) \)[/tex]:
[tex]\[ y - (-4a) = -2(x - a) \][/tex]
[tex]\[ y + 4a = -2x + 2a \][/tex]
[tex]\[ y = -2x + 2a - 4a \][/tex]
[tex]\[ y = -2x - 2a \][/tex]
So, the equation of the tangent line is:
[tex]\[ y = -2x - 2a \][/tex]
### Step 4: Find the Slope of the Normal Line
The normal line is perpendicular to the tangent line. The slope of the normal line is the negative reciprocal of the slope of the tangent line.
Since the slope of the tangent is [tex]\(-2\)[/tex], the slope of the normal line is:
[tex]\[ m_{\text{normal}} = \frac{1}{2} \][/tex]
### Step 5: Find the Equation of the Normal Line
Using the point-slope form of the equation of a line again:
[tex]\[ y - (-4a) = \frac{1}{2}(x - a) \][/tex]
[tex]\[ y + 4a = \frac{1}{2}x - \frac{1}{2}a \][/tex]
[tex]\[ 2(y + 4a) = x - a \][/tex]
[tex]\[ 2y + 8a = x - a \][/tex]
[tex]\[ 2y = x - 9a \][/tex]
[tex]\[ x - 2y = 9a \][/tex]
So, the equation of the normal line is:
[tex]\[ x - 2y = 9a \][/tex]
### Summary
- The equation of the tangent to the parabola [tex]\( y^2 = 16ax \)[/tex] at the point where the ordinate is [tex]\(-4a\)[/tex] is:
[tex]\[ y = -2x - 2a \][/tex]
- The equation of the normal to the same parabola at the same point is:
[tex]\[ x - 2y = 9a \][/tex]
