Locus + other Q's (1 Viewer)

[-pari-]

with exams next week, i've actually decided to study....
any help muchly appreciated


1) Equation of a parabola with vertex (3, -2) focus (7, -2) is y^2 + 4y - 16x + 52 = 0.
Find the equation of the tangent to the parabola at the point where x = 4 in the first quadrant

first - how on earth do i find the first derivative of that equation?
second.....whats with the "in the first quadrant bit"

answer: 2x - y - 6 = 0.

2) find the equation of the locus of midpoints of all chords of the length 2units in the circle with equation x^2 + y^2 - 2y - 3 = 0.

no idea how to deal with this one.....answer: x^2 + y^2 - 2y - 2 = 0.

3) Find the equation of the parabola with axis parallel to the y axis and passing through points (0, -2) (1, 0) (3, -8)

Answer: y = -2x^2 + 4x - 2

4) the point P(x, y ) lies on the parabola y = (1/2)x^2
Find this point such that the sum of the abscissa an ordinate is a ninimum.
Hint: Sum = x + (1/2)x^2

(how'd they get the equation for the sum?)

Answer: P (-1, 1/2)

 

[SoulSearcher]

Well y² + 4y +52 = 16x
y² + 4y + 4 = 16x - 48
(y+2)² = 16(x-3)
(y+2)²/16 = x-3
(y+2)²/16 + 3 = x
dx/dy = (y+2)/8
dy/dx = 8/(y+2)

That's the derivative. The second bit is that this parabola is of the form x = (y-k)², so it is a parabola that for every x-value in its domain, it has two corresponding y-values, so you would choose the y-value of the graph when x = 4 that is positive, since it is in the first quadrant.

Second question, you find the centre of the circle, which can be done by completing the square. The equation should be x² + (y-1)² = 2
You now realise the centre of the circle is (0, 1) and the radius of this circle is rt2 units.
Knowing this, and also knowing a theorem that the radius of a circle bisects a chord of the circle at right angles, and since the length of the chord is 2 units, then you have a right-angled triangle inscribe within the circle, the hypotenuse being rt2 units, one of the sides being 1 unit in length (radius bisects the chord), and then you can find the radius of the other circle, which will be 1 unit in length. Therefore the equation of that circle would be x² + (y-1)² = 1, which on expansion will become x² + y² - 2y - 2 = 0

Still working on the third question, will get back to you on that.

 

[SoulSearcher]

Fourth question, the sum is of the abscissa and the ordinate, which would be the x value of the abscissa, I think that's how they got their sum, wouldn't have a clue. From there, differentiatie, then find x and y co-ordinates of point P.

 

[Mattamz]

For question 3:
Let the equ. of the parabola being y = ax^2 + bx + c.

Sub (0,-2):
-2 = 0 + 0 + c
c = -2

Sub (1,0):
0 = a + b + c
0 = a + b - 2 (since c=-2)
a + b = 2

Sub (3, -8):
-8 = 9a + 3b -2
-6 = 6a + 3a +3b
-6 = 6a + 3(a +b)
-6 = 6a + 6 (since a + b = 2)
6a = -12
a = -2

Sub a= -2:
-2 + b = 2
b = 4

therefore the equ. of the parabola is y = -2x^2 +4x -2

 

[-pari-]

^ damn. that was pretty easy. thanks =)
a few more...

1) a cable care 100m above the ground is seen to have an angle of elevation of 65degrees when it is on a bearing of 345degrees. after a minute it has an angle of elevation of 69degrees and is on a bearing of 025degrees
Find how far it travels in that minute and its speed in ms^(-1)
answer: 30.1m 0.5ms^(-1)

a ball is dropped from a height of 12m. on the first rebound it rises to 10m. at subsequent rebounds it rises to a height equal to 5/6 of that which it previously attained
when the ball is dropped it takes 8 seconds to reach the floor. on the first rebound it takes 6 sec. to reach the floor. at subsequent rebounds it takes 3/4 of the time taken for the previous rebound.
calculate the total time before the balls comes to a rest.

answer: 32sec

 

[carolineedwards]

I don't think question two was answered quite right.

The idea with the inscribe right angle triangle is right but the hypotenuse is 2, the halved side with the chord is 1 and the side that forms the right angle with the chord is root 3 due to pythag theorem. So the equation of the circle that joins the mid points is actually x^2+(y-1)^2=3 and that expanded out is x^2+y^2-2y-2=0 i.e the answer. You did your expanding wrong --> your answer should have been; x^2+y^2-2y=0.

 

[jathu123]

lol 9 years ago, I don't think they'll read it haha