I'm having difficulties with working out the answer for this:
Find the range of values of k for which y= x-3 meets x^2 - 3y^2= k in two distinct points,
Please help & show the methods if you could! Thank you:)
Find the range of values of k for which y= x-3 meets x^2 - 3y^2= k in two distinct points.?
2010-05-13 9:28 pm
回答 (5)
2010-05-13 9:39 pm
✔ 最佳答案
plug y for interceptx^2 -3(x-3)^2 = k
x^2 -3x^2 +18x -27=k
2x^2 - 18x +(k+27) =0
x= (18+- sqrt( 18^2-8(k+27) )/4
For two distinct points , two distinct roots , for two distinct roots , 18^2 +8(k+27) >0
( Discriminant >0)
324 +8 k +216 >0
8k > -540
k> -135/2
HI ED !! A GREETING FROM CHILE
2016-11-04 11:19 pm
you're able to be able to desire to discover the place they intersect, with the aid of substituting the 2d equation into the 1st. this supply you some algebra below a sq. root sign that could purely be solved with the aid of inequality. because of the fact the equation has no actual roots, then the equation could desire to be < 0.
2010-05-13 9:39 pm
If y = x - 3 then:
y^2 = x^2 - 6x + 9
3y^2 = 3x^2 - 18x + 27
If x^2 - 3y^2 = k then:
-3y^2 = k - x^2
3y^2 = x^2 - k
If both are true, then we have:
3x^2 - 18x + 27 = x^2 - k
3x^2 - 18x + 27 - x^2 + k = 0
2x^2 - 18x + (27 + k) = 0
For two distinct solutions, we need the discriminant to be positive, i.e.:
(-18)^2 - 4(2)(27 + k) > 0
324 - 216 - 8k > 0
108 - 8k > 0
108 > 8k
k < 108 / 8
k < 13.5
y^2 = x^2 - 6x + 9
3y^2 = 3x^2 - 18x + 27
If x^2 - 3y^2 = k then:
-3y^2 = k - x^2
3y^2 = x^2 - k
If both are true, then we have:
3x^2 - 18x + 27 = x^2 - k
3x^2 - 18x + 27 - x^2 + k = 0
2x^2 - 18x + (27 + k) = 0
For two distinct solutions, we need the discriminant to be positive, i.e.:
(-18)^2 - 4(2)(27 + k) > 0
324 - 216 - 8k > 0
108 - 8k > 0
108 > 8k
k < 108 / 8
k < 13.5
2010-05-13 9:37 pm
x^2 - 3y^2 = k
x^2 - 3(x-3)^2 = k
x^2 - 3(x^2 -6x +9) = k
-2x^2 + 18x - 27 = k
2x^2 + 18x -(k+27) = 0
x^2 + 9x -(k+27)/2 = 0
x = (9 +- sqrt((81 + 4(k+27))/2)/2 = 0
x = (3 +- sqrt(2k+135))/2
to have 2 real solutions, the value inside sqrt must be positive. 2k+135 > 0
k > -67.5
x^2 - 3(x-3)^2 = k
x^2 - 3(x^2 -6x +9) = k
-2x^2 + 18x - 27 = k
2x^2 + 18x -(k+27) = 0
x^2 + 9x -(k+27)/2 = 0
x = (9 +- sqrt((81 + 4(k+27))/2)/2 = 0
x = (3 +- sqrt(2k+135))/2
to have 2 real solutions, the value inside sqrt must be positive. 2k+135 > 0
k > -67.5
2010-05-13 9:37 pm
y = x - 3
x^2 - 3(x - 3)^2 = k
x^2 - 3(x^2 - 6x + 9) = k
x^2 - 3x^2 + 18x - 27 = k
-2x^2 + 18x - 27 = k
-2x^2 + 18x - (k + 27) = 0
For two solutions, b^2 - 4ac = (18)^2 - 4(-2)(-k - 27) > 0
324 + 8k + 216 > 0
8k > -540
k > 67.5
x^2 - 3(x - 3)^2 = k
x^2 - 3(x^2 - 6x + 9) = k
x^2 - 3x^2 + 18x - 27 = k
-2x^2 + 18x - 27 = k
-2x^2 + 18x - (k + 27) = 0
For two solutions, b^2 - 4ac = (18)^2 - 4(-2)(-k - 27) > 0
324 + 8k + 216 > 0
8k > -540
k > 67.5
收錄日期: 2021-05-01 12:05:59
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