2002 Winter diet paper 2 Q.10
Posted 19 May 2005 - 02:11 PM
Posted 19 May 2005 - 02:38 PM
Re-arrange the equation of the line so you have y= k - 2x.
Substitute this value of y into the equation for the circle.
A line is a tangent to a circle where b - 4ac = 0, so you want to arrange the above equation ang get the following:
5x - (2+4k)x + (k -4) = 0
Now using the discriminant, b - 4ac = 0, and factorising you should get that k = 7 OR k = -3 and since you are asked to find the value of k where k > 0 you can conclude that the value for k is k = 7.
That make sense???
Posted 19 May 2005 - 02:46 PM
At this piont, suppose your equation was:
5x sq - (1 + 2k) 2x ....
Wouldn't it make more sense since you've taken out the common factor of 2x, not just x ? Or does the coefficient of x outside the bracket ALWAYS have to be 1?
Posted 19 May 2005 - 02:51 PM
ax + bx + c
If you try to calculate the discriminant of a quadratic you must have in the above form.
a is the coefficient of x
b is the coefficient of x
c is a constant term
Therefore, your discriminant value would be incorrect as it is the coefficient of the x term on its own.
That make sense??
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