PN is an ordinate of the parabola y2= 4ax. A straight line is drawn pa...
The equation of parabola be y2=4ax
let the point P be (at2,2at)
PN is ordinate ⇒N(at2,0)
Equation of straight line bisecting NP is
y=at
substituting y in equation of parabola
a2t2 = 4ax
⇒x = 4at2
So the coordinates of Q are (4at2 ,at)
Equation of NQ is y−0 = (at−0)/(at2/4 - at2)(x−at2)
y= −4/3t(x−at2)
Put x=0
y = −4/3t(0−at^2)
⇒y=4at/3
⇒AT = 4at/3
NP = 2at
AT/NP = (4at/3)/2at
= ⅔
AT = 2/3NP
PN is an ordinate of the parabola y2= 4ax. A straight line is drawn pa...
Given:
- Parabola equation: y^2 = 4ax
- PN is an ordinate of the parabola
- A straight line is drawn parallel to the axis to bisect NP and meets the curve in Q
- NQ meets the tangent at the vertex in a point T such that AT = kNP
- A is the vertex
To find: The value of k
Solution:
1. Drawing the diagram
- Let's draw a diagram to better visualize the problem.
- We know that PN is an ordinate of the parabola, which means it is perpendicular to the x-axis.
- Also, a line is drawn parallel to the axis of the parabola to bisect NP. Let's call this line MN.
- MN intersects the parabola at point Q.
- NQ meets the tangent at the vertex, which means it passes through the vertex. Let's call this point of intersection T.
- We need to find the value of k, given that AT = kNP.
- Here's a rough sketch of the diagram:
T
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V (vertex)
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Q |
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N |
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P * * * * * * * |
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2. Finding the coordinates of P and N
- Since PN is perpendicular to the x-axis, we know that its x-coordinate is the same as that of Q. Let's call this x-coordinate a.
- We also know that PN is bisected by line MN. Let's call the midpoint of PN as M.
- Since M is the midpoint of PN, its y-coordinate is the average of the y-coordinates of P and N. Let's call the y-coordinate of M as b.
- Therefore, the coordinates of P are (a, 2b) and the coordinates of N are (a, 0).
3. Finding the coordinates of Q
- We know that Q lies on the parabola y^2 = 4ax and also on the line MN.
- Since MN is parallel to the x-axis, we know that the y-coordinate of Q is b.
- To find the x-coordinate of Q, we can substitute y = b in the equation of the parabola:
b^2 = 4ax
a = b^2/4x
Putting this value of a in x = a, we get:
x = b^2/4a
- Therefore, the coordinates of Q are (b^2/4a, b).
4. Finding the equation of the tangent at the vertex
- The vertex of the parabola is at (0, 0).
- The equation of the parabola is y^2 = 4ax.
- To find the equation of the tangent at the vertex, we need to find the derivative of y with respect to x:
dy/dx = (1/2)(4ax)^(-1/2) * 4a
= 2a^(-1/2)
- At the vertex (0, 0), the derivative is infinite. Therefore, the equation of the tangent at the vertex is
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