Soal SBMPTN 2018 + Pembahasan No. 10

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10. Jika garis singgung kurva y = 9 – x2 di titik P(a, b) dengan b > 0 memotong sumbu x di titik Q(-5, 0), maka ab adalah...
(A) -10 
(B) -8 
(C) 0 
(D) 8
(E) 10


Jawaban: B
P(a,b) paday=9-x2
b = 9 - a2

gradien m = y' ® m = -2x di titik (a, b)
m = -2a P.G.S® y-b=-2a(x-a)
y - (9 - a2) = -2a(x - a)
melalui Q(-5, 0) ® -9 + a2 = -2a(-5 - a)

-9 + a2 = 10a + 2a2 0 = a2 + 10a + 9
0 = (a + 1)(a + 9) a=-1 Ú a=-9 b = 8 b = -55 b>0 (TM)
jadi a · b = -8

Soal SBMPTN 2018 + Pembahasan No. 10 Soal SBMPTN 2018 + Pembahasan No. 10 Reviewed by Satuan Kejuruan on April 20, 2019 Rating: 5

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