parabola help please (1 Viewer)

[clintmyster]

Hi i have these questions to do and im kinda lost..help would be much appreciated. For the last question, i only need the solution for the proving part, the rest i did. Thanks

 

[independantz]

10)
i) for this q differentiate y, then let itequal 2, since the line is parallel to the tangent, they have the same gradient. That will get you the x-coordinate of the point of contact sub it back into the original eq to get the y coord.

ii)basically do the same as in i) but when you let the derivate = the gradient , reciprocate and negate the gradient, so it becomes a normal. The rest is just the same.

11) iii) Find the directrix, y=-a, generally, then just equate it to the eq of the circle you found in ii) , it should only have one solution, which will be the point of contact.

 

[clintmyster]

10)
i) for this q differentiate y, then let itequal 2, since the line is parallel to the tangent, they have the same gradient. That will get you the x-coordinate of the point of contact sub it back into the original eq to get the y coord.

ii)basically do the same as in i) but when you let the derivate = the gradient , reciprocate and negate the gradient, so it becomes a normal. The rest is just the same.

11) iii) Find the directrix, y=-a, generally, then just equate it to the eq of the circle you found in ii) , it should only have one solution, which will be the point of contact.

I got 11 okay, i thought it was like that however with 10 still a bit confused. Mind positing up the solution? I always have issues differentiating so itd be quite helpful. Thanks

 

[vds700]

I got 11 okay, i thought it was like that however with 10 still a bit confused. Mind positing up the solution? I always have issues differentiating so itd be quite helpful. Thanks

i) y^2 = 8x
y = sqrt(8x) we use the chain rule, let u = 8x
y = u^1/2
dy/dx = dy/du . duy/dx
=1/2 u^-1/2 . 8
=4/u^1/2
=4/(8x)^1/2

this gives us the gradient of the tangent at any point x on the upper half of the parabola

we require the gradient of the tangent to be 2

4/(8x)^1/2 = 2 ... square both sides
16/8x = 4
32x = 16
x = 1/2

when x = 1/2
y = sqrt 4 = 2

therefore the point of contact of this tangent is (1/2, 2)

ii) 2x + 3y + 5 = 0
3y = -2x-5
y = (-2x/3) - 5/3

required gradient of the normal is -2/3, therefore the gradient of the tanfent at this point is 3/2

y^2 = 12x
y = (12x)^1/2
ill let u differentiate this, same way as i did above, make that equal to 3/2 to find the x-coordinate of the point of contact of the normal, then sub this into y =(12x)^1/2 to find thw y-coordinate

 

[clintmyster]

thanks vds700 this helped heaps!