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#1
find eq/n of line that is a tangent to both y^2 = 4ax and x^2 = 4ay answer is x + y + a = 0, can someone show me how pls?

and also
y^2 = 4x, find the eq/n of the normal to that that goes thru the point
(21/4,3/2)

a thankyou

G
0
14 years ago
#2
Please help- i'm stuck on the same question too! Its on page 79 of heinemman p5 book. do we need to use parametric coordinates?
Thanks
0
14 years ago
#3
For y²=4ax at (at², 2at), plugging the coordinates into the tangent equation:
ty = at² + x
y = at + (x/t)

Subbing into x² = 4ay:
x² = 4a²t + (4a/t)x
x² - (4a/t)x - 4a²t = 0
As the tangent only touches the curve once this quadratic can only have one solution for x so the b²-4ac bit of the quadratic formula is 0:
16a²/t² + 16a²t = 0
16a² + 16a²t³ = 0
t=-1

Eqtn of tangent:
ty = at² + x
-y = a + x.
x + y + a = 0
0
14 years ago
#4
y^2 = 4x, find the eq/n of the normal to that that goes thru the point
(21/4,3/2)

Equation of normal of y² = 4ax at (at², 2at) is:
y + tx = 2at + at³
In this case a=1 and we know that (x,y) = (21/4 , 3/2) satisfies the equation. So subbing all that in:
(3/2) + (21/4)t = 2t + t³
6 + 21t = 8t + 4t³
4t³ - 13t - 6 = 0
(t - 2)(2t + 1)(2t + 3) = 0
So the normal will pass through (21/4, 3/2) when either t=2, t=-(1/2) or t=-(3/2)
Sub these into y + tx = 2t + t³ to get three possible equations.
0
#5
ahhh thanks, i thought the t values would be different on each curve 'cause it gives diff y and x values.

also i got that t^3 eq/n but stopped becos i thought i had gone wrong, is there an easy way u use to just bang them into 3 brackets like that? cheers

G
0
14 years ago
#6
Thankyou
0
14 years ago
#7
np
For the 3 brackets. I just factorized it on the computer because I couldn't be bothered. The full working of that factorization isn't important in the solution but in an exam you'd try values of t until the thing was zero. Then you'd use the factor theorem and divide to get an easily solvable quadratic.
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