Parabola questions (1 Viewer)

[reggie7]

1. Find the equation of the normal to the curve x^2 = 12y at the point (6 , 3).
The normal meets the parabola again at point P. Find the coordinates of P.

2. The normal of the parabola x^2 = 18y at (-6 , 2) cuts the parabola again at Q. Find the coordinates of Q.


3. Find the equations of the normals to the curve x^2 = -8y at the points (-16 , -32) and (-2 , -0.5). Find their point of intersection and show that this point lies on the parabola.
please explain the working out:wave:

 

[Zeber]

12y = x^2
y = x^2/6
dy/dx = 2x/6 = x/3 x = 6 at point(6,3)

dy/dx of tangent at given point = 6/3 = 2, but we want normal so dy/dx = -1/2

y - 3 = -1/2 *(x-6)
2y - 6 = -x + 6
2y + x - 12 = 0

since y = x^2/12

sub this into the equation of normal and solve.

 

[gurmies]

1.

12y = x^2

y = x^2/12

dy/dx = x/6

At (6,3) Slope of tangent is 1. Therefore slope of normal is -1.

y - 3 = - (x - 6)

y - 3 = -x + 6

x + y - 9 = 0

Now, it meets again at point "P". Therefore:

x + (x^2/12) - 9 = 0 (I subbed in y = x^2/12)

x^2 + 12x - 108 = 0

(x-6)(x+18) = 0

x = 6 or/ x = -18

Now we already know that x = 6 is a point of intersection, therefore the x values we are looking for is -18. Subbing into either the equation of the parabola or the equation of the normal, we get that y = 27. Therefore the co-ordinates of "P" are (-18,27)


2.

18y = x^2

y = x^2/18

dy/dx = x/9

At (-6, 2) Slope of tangent is -2/3. Therefore slope of normal is 3/2.

y - 2 = 3/2(x + 6)

2y - 4 = 3x + 18

3x - 2y + 22 = 0

And we know that y = x^2/18 (from equation of parabola)

3x - 2(x^2/18) + 22 = 0

3x - (x^2/9) + 22 = 0

27x - x^2 + 198 = 0

x^2 - 27x - 198 = 0

(x+6)(x-33) = 0

x = -6 or/ x = 33 (we already know that x = -6 is a solution)

Therefore the x-value we are looking for is 33. When x = 33, y = 60.5. Therefore co-ordinates are (33, 60.5)

3.

x^2 = -8y

y = x^2/-8

dy/dx = -x/4

At (-16, -32) Slope of Tangent is 4. Therefore Slope of Normal is -1/4

y + 32 = -1/4(x + 16)

4y + 128 = -x - 16

x + 4y + 144 = 0 (1)

At (-2, -0.5) Slope of Tangent is 1/2. Therefore Slope of Normal is -2.

y + 0.5 = -2 (x + 2)

y + 0.5 = -2x - 4

2x + y + 4.5 = 0

4x + 2y + 9 = 0 (2)

Solving (1) and (2) simultaneously:

x + 4y + 144 = 0 (1)

4x + 2y + 9 = 0 (2)

(2) x 2 = 8x + 4y + 18 = 0 (3)

(3) - (1) 7x -126 = 0

7x = 126

x = 18

Therefore, y = -40.5

Therefore, point of intersection is (18, -40.5).

Testing on x^2 = -8y

LHS = (18)^2 = 324

RHS = -8(-40.5) = 324

LHS = RHS, and thus the intersection of these two normals also lies on the parabola x^2 = -8y

 

[bored of sc]

Solid post gurmies.

 

[gurmies]

thanks man =) By helping others, I'm also rekindling & testing my own knowledge!