Final answer:
To find the length of the perpendicular from the focus of the parabola onto the normal, we first need to find the equation of the normal. By differentiating the equation of the parabola and finding the slope of the tangent, we can determine the slope of the normal. Then, we can use the distance formula to find the length of the perpendicular line. Finally, substituting the value of the perpendicular length into the given equation gives the answer as A. 1.
Step-by-step explanation:
To find the length of the perpendicular from the focus of the parabola onto the normal, we need to first find the equation of the normal. The given equation of the parabola is y²=12x, which is in the standard form of a parabola. The normal to the parabola at any point (x, y) is perpendicular to the tangent at that point.
The slope of the tangent at any point on the parabola can be found by differentiating the equation of the parabola with respect to x. After finding the slope of the tangent, we can find the slope of the normal by taking the negative reciprocal of the tangent slope.
Let's differentiate the equation of the parabola: 2y(dy/dx) = 12. Simplifying the equation gives dy/dx = 6/y.
Therefore, the slope of the tangent at any point is 6/y. The slope of the normal is the negative reciprocal of the tangent slope, which is -y/6.
We are given that the equation of the normal is x+y=k. Substituting the slope of the normal, -y/6, into the equation of the line gives x - y/6 = k.
To find the length of the perpendicular from the focus of the parabola onto the normal, we need to find the distance between these two lines. The formula for the distance between a point and a line is given by:
d = |Ax + By + C| / sqrt(A² + B²),
where the line is Ax + By + C = 0, and (x, y) is any point on the line.
In this case, we have two lines x+y=k and y²=12x, and the focus of the parabola can be found using the equation x = a, where a is the distance from the focus to the vertex of the parabola.
We know that for a parabola, the distance from the vertex to the focus is equal to a, where a is the constant in the equation y²=4ax.
Therefore, the distance from the focus to the vertex is a/2.
In this case, the equation of the parabola is y²=12x, so a = 12 and the distance from the focus to the vertex is 6.
Therefore, the focus is at (6, 0).
Substituting the values into the distance formula gives d = |6 - k/7 + 0| / sqrt(1² + 1/7²) = |6 - k/7| / sqrt(1 + 1/49) = |6 - k/7| / sqrt(50/49) = |6 - k/7| * sqrt(49/50) = |42 - k/7| / sqrt(50).
Finally, we can substitute the value of p, which is the length of the perpendicular from the focus to the normal, into the given equation 4k - 2p².
Since p is the distance between the focus and the normal, we can replace p with |42 - k/7| / sqrt(50), giving 4k - 2(|42 - k/7| / sqrt(50))².
Simplifying further gives:
4k - 2(|42 - k/7| / sqrt(50))² = 4k - 2(|42 - k/7|² / 50).
Therefore, the answer is A. 1.