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The area of the triangle formed by the tangent and the normal to the parabola y2 = 4ax, both drawn at the same end of the latus rectum and the axis of the parabola, is
- a)2√2a2
- b)2a2
- c)4a2
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
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The area of the triangle formed by the tangent and the normal to the p...
Let's start by finding the equation of the tangent and the normal to the parabola.
The equation of the parabola is y^2 = 4ax.
To find the slope of the tangent at a point on the parabola, we differentiate the equation of the parabola with respect to x:
2yy' = 4a
y' = 2a/y
Let's consider a point P (x1, y1) on the parabola.
The equation of the tangent at point P is y - y1 = y'(x - x1):
y - y1 = 2a/y1 * (x - x1)
y - y1 = (2ax - 2ax1) / y1
yy1 - y1^2 = 2ax - 2ax1
yy1 = 2ax + 2ax1 - y1^2
Similarly, the equation of the normal at point P is given by:
y - y1 = -y'(x - x1)
y - y1 = -2a/y1 * (x - x1)
yy1 - y1^2 = -2ax + 2ax1
yy1 = -2ax + 2ax1 + y1^2
Now, let's find the coordinates of the points where the tangent and the normal intersect the axis of the parabola.
The equation of the axis of the parabola is y = 0.
For the tangent:
yy1 = 2ax + 2ax1 - y1^2
0 = 2ax + 2ax1 - y1^2
2ax + 2ax1 = y1^2
x + x1 = y1^2 / (2a)
For the normal:
yy1 = -2ax + 2ax1 + y1^2
0 = -2ax + 2ax1 + y1^2
2ax - 2ax1 = y1^2
x - x1 = y1^2 / (2a)
So, the x-coordinate of the point where the tangent intersects the axis of the parabola is x + x1 = y1^2 / (2a), and the x-coordinate of the point where the normal intersects the axis of the parabola is x - x1 = y1^2 / (2a).
The distance between these two points is 2(x - x1) = 2(y1^2 / (2a)) = y1^2 / a.
The height of the triangle formed by the tangent and the normal is y1.
Therefore, the area of the triangle is (1/2) * base * height = (1/2) * (y1^2 / a) * y1 = (y1^3) / (2a).
So, the area of the triangle formed by the tangent and the normal to the parabola y^2 = 4ax, both drawn at the same end of the latus rectum and the axis of the parabola, is (y1^3) / (2a).
Therefore, the correct option is a) 2.
The equation of the parabola is y^2 = 4ax.
To find the slope of the tangent at a point on the parabola, we differentiate the equation of the parabola with respect to x:
2yy' = 4a
y' = 2a/y
Let's consider a point P (x1, y1) on the parabola.
The equation of the tangent at point P is y - y1 = y'(x - x1):
y - y1 = 2a/y1 * (x - x1)
y - y1 = (2ax - 2ax1) / y1
yy1 - y1^2 = 2ax - 2ax1
yy1 = 2ax + 2ax1 - y1^2
Similarly, the equation of the normal at point P is given by:
y - y1 = -y'(x - x1)
y - y1 = -2a/y1 * (x - x1)
yy1 - y1^2 = -2ax + 2ax1
yy1 = -2ax + 2ax1 + y1^2
Now, let's find the coordinates of the points where the tangent and the normal intersect the axis of the parabola.
The equation of the axis of the parabola is y = 0.
For the tangent:
yy1 = 2ax + 2ax1 - y1^2
0 = 2ax + 2ax1 - y1^2
2ax + 2ax1 = y1^2
x + x1 = y1^2 / (2a)
For the normal:
yy1 = -2ax + 2ax1 + y1^2
0 = -2ax + 2ax1 + y1^2
2ax - 2ax1 = y1^2
x - x1 = y1^2 / (2a)
So, the x-coordinate of the point where the tangent intersects the axis of the parabola is x + x1 = y1^2 / (2a), and the x-coordinate of the point where the normal intersects the axis of the parabola is x - x1 = y1^2 / (2a).
The distance between these two points is 2(x - x1) = 2(y1^2 / (2a)) = y1^2 / a.
The height of the triangle formed by the tangent and the normal is y1.
Therefore, the area of the triangle is (1/2) * base * height = (1/2) * (y1^2 / a) * y1 = (y1^3) / (2a).
So, the area of the triangle formed by the tangent and the normal to the parabola y^2 = 4ax, both drawn at the same end of the latus rectum and the axis of the parabola, is (y1^3) / (2a).
Therefore, the correct option is a) 2.
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