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If the three normals drawn to the parabola y2 = 2x pass through the point (a, 0), a ≠ 0, then 'a' must be greater than:
- a)1/2
- b)-1/2
- c)-1
- d)1
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
If the three normals drawn to the parabola y2 = 2x pass through the po...
To find the equation of the normal to the parabola at a given point, we need to find the slope of the tangent at that point and then take the negative reciprocal of it.
The equation of the parabola is y^2 = 2x. We can rewrite it as y = √(2x) or y = -√(2x).
To find the slope of the tangent at a point (x, y), we can use the derivative. Taking the derivative of y = √(2x), we get:
dy/dx = 1/√(2x)
To find the slope of the tangent at a point (x, y), we need to substitute x into the derivative:
m = 1/√(2x)
Now, let's find the equation of the normal passing through the point (a, 0). We know that the slope of the normal is the negative reciprocal of the slope of the tangent:
m_normal = -√(2a)
Using the point-slope form of the equation of a line, we can write the equation of the normal as:
(y - 0) = -√(2a)(x - a)
Simplifying, we get:
y = -√(2a)x + √(2a^3)
Therefore, the equation of the normal passing through the point (a, 0) is y = -√(2a)x + √(2a^3).
The equation of the parabola is y^2 = 2x. We can rewrite it as y = √(2x) or y = -√(2x).
To find the slope of the tangent at a point (x, y), we can use the derivative. Taking the derivative of y = √(2x), we get:
dy/dx = 1/√(2x)
To find the slope of the tangent at a point (x, y), we need to substitute x into the derivative:
m = 1/√(2x)
Now, let's find the equation of the normal passing through the point (a, 0). We know that the slope of the normal is the negative reciprocal of the slope of the tangent:
m_normal = -√(2a)
Using the point-slope form of the equation of a line, we can write the equation of the normal as:
(y - 0) = -√(2a)(x - a)
Simplifying, we get:
y = -√(2a)x + √(2a^3)
Therefore, the equation of the normal passing through the point (a, 0) is y = -√(2a)x + √(2a^3).
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Community Answer
If the three normals drawn to the parabola y2 = 2x pass through the po...
Let the equation of the normal be:
⇒ y = mx − 2am −am3
Here, 4a = 2
⇒ a = 1/2
⇒ y = mx − m − (1/2)m3
It passing through A(a, 0).
Then 0 = am −m − (1/2)m3
⇒ m = 0, a − 1 − (1/2)m2 = 0
⇒ m2 = 2(a − 1) > 0
⇒ a > 1
⇒ y = mx − 2am −am3
Here, 4a = 2
⇒ a = 1/2
⇒ y = mx − m − (1/2)m3
It passing through A(a, 0).
Then 0 = am −m − (1/2)m3
⇒ m = 0, a − 1 − (1/2)m2 = 0
⇒ m2 = 2(a − 1) > 0
⇒ a > 1
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If the three normals drawn to the parabola y2 = 2x pass through the point (a, 0), a ≠ 0, then a must be greater than:a)1/2b)-1/2c)-1d)1Correct answer is option 'D'. Can you explain this answer?
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