A Maths CE 06 Q14
2011-03-08 12:13 am
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Let J be the circle x^2 + y^2 = r^2, where r > 0.a) Suppose that the straight line L: y = mx + c
is a tangent to J.
i) Show that c^2 = r^2 ( m^2 + 1).
ii) If L passes through a point (h, k), show that
(k - mh)^2 = r^2 (m^2 + 1).b) J is inscribed in a triangle PQR ( see the figure). The coordinates of P and R are (7, 4)
and (-5, -5) respectively.
i) Find the radius of J.
ii) Using (a)(ii), or otherwise, find the slope of
PQ.
iii) Find the coordinates of Q.
回答 (2)
2011-03-08 12:48 am
✔ 最佳答案
(a)(i) Sub. y = mx + c into J: x^2 + y^2 = r^2x^2 + (mx + c)^2 = r^2
(1 + m^2)x^2 + 2mcx + (c^2 - r^2) = 0
Discriminant = 0
4m^2c^2 - 4(1 + m^2)(c^2 - r^2) = 0
m^2c^2 - c^2 + r^2 - m^2c^2 + m^2r^2 = 0
c^2 = r^2(m^2 + 1)
(ii) L passes through (h,k) => k = mh + c
(k - mh)^2 = r^2(m^2 + 1)
(b)(i) P(7,4) R(-5,-5). m = 9/12 = 3/4
Sub. P and m into (k - mh)^2 = r^2(m^2 + 1)
(4 - 21/4)^2 = r^2(9/16 + 1)
25/16 = r^2(25/16)
r = 1
(ii) Sub. P and r = 1 into (k - mh)^2 = r^2(m^2 + 1)
(4 - 7m)^2 = (m^2 + 1)
48m^2 - 56m + 15 = 0
(4m - 3)(12m - 5) = 0
m = 3/4 or 5/12
The slope of PQ is 5/12
(iii) Sub. R and r = 1 into (k - mh)^2 = r^2(m^2 + 1)
(-5 + 5m)^2 = (m^2 + 1)
12m^2 - 25m + 12 = 0
(4m - 3)(3m - 4) = 0
m = 3/4 or 4/3
The slope of PR is 4/3
Let Q(s,t) Then (t - 4)/(s - 7) = 5/12 and (t + 5)/(s + 5) = 4/3
5s = 12t - 13
4s = 3t - 5
s = -7/11 and t = 9/11. Q(-7/11,9/11)
2011-03-08 6:19 am
biii ) 有唔明...
4s = 3t - 5
跟住點計到 s = -7/11 and t = 9/11
4s = 3t - 5
跟住點計到 s = -7/11 and t = 9/11
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