***20. Show that the length of the subnormal at any point on the curve y2 = 4ax is a constant.
Answers
Step-by-step explanation:
The length of the subnormal is always constant and is equal to the semi-latus rectum i.e. 2a …Hence D is the correct. This discussion on The length of the sub normal to the parabola y2=4ax at any point is equal toa)√2ab)2√2ac)a/√2d)2aCorrect answer is option 'D'.
Answer:
The length of the subnormal is a constant, independent of the point (x1, y1) on the curve.
Step-by-step explanation:
To find the length of the subnormal at any point on the curve y^2 = 4ax, we first need to find the equation of the tangent and then use the definition of the subnormal.
Let (x1, y1) be a point on the curve y^2 = 4ax. We can find the equation of the tangent to the curve at this point by differentiating the equation of the curve with respect to x and evaluating it at x1:
dy/dx = 2a/y
At the point (x1, y1), the slope of the tangent is:
dy/dx = 2a/y1
The equation of the tangent is therefore:
y - y1 = 2a/y1 (x - x1)
We can now use the definition of the subnormal to find its length. The subnormal is the line segment that is perpendicular to the tangent and intersects the x-axis. Let (x2, 0) be the point where the subnormal intersects the x-axis. Since the subnormal is perpendicular to the tangent, its slope is the negative reciprocal of the slope of the tangent:
m_subnormal = -y1/2a
The equation of the subnormal is therefore:
y = -y1/2a x + y1
To find the point (x2, 0) where the subnormal intersects the x-axis, we set y = 0 in the equation of the subnormal:
0 = -y1/2a x2 + y1
x2 = 2a
The length of the subnormal is the distance between the point (x1, y1) and the point (2a, 0):
length = sqrt((2a - x1)^2 + y1^2)
We can simplify this expression using the equation of the curve y^2 = 4ax:
length = sqrt((2a - x1)^2 + 4a^2/x1)
To show that this expression is a constant, we can take its derivative with respect to x1 and show that it is equal to zero:
d/dx1(length^2) = 2(2a - x1)(-1) + 4a^2(-1/x1^2)(2a - x1)
Simplifying this expression, we get:
d/x1(length^2) = -2(2a - x1) + 8a^2(2a - x1)/x1^2
d/dx1(length^2) = -2(2a - x1)(1 - 4a/x1)
At the point (x1, y1), we have x1 = y1^2/4a. Substituting this into the expression above, we get:
d/dx1(length^2) = -2(y1^2/2a)(1 - 1) = 0
Therefore, the length of the subnormal is a constant, independent of the point (x1, y1) on the curve.
To learn more about similar questions visit:
https://brainly.in/question/6558024?referrer=searchResults
https://brainly.in/question/5334519?referrer=searchResults
#SPJ6