Question 27.

Let a, b, x, y be real numbers such that a2+b2=25,x2+y2=169a^2 + b^2 = 25, x^2 + y^2 = 169, and ax+by=65ax + by = 65. If k=aybxk = ay - bx, then

A
0<k5130 \lt k \leq \frac{5}{13}
B
k>513k \gt \frac{5}{13}
C
k=513k = \frac{5}{13}
D
k = 0

Question Explanation

Text Explanation

(ax+by)2=652\left(ax+by\right)^2=65^2

a2x2 + b2y2+ 2abxy = 652a^2x^2\ +\ b^2y^2+\ 2abxy\ =\ 65^2

k=aybxk = ay - bx

k2 = a2y2+b2x22abxyk^2\ =\ a^2y^2+b^2x^2-2abxy

(a2+b2)(x2+y2)=25169(a^2 + b^2)(x^2 + y^2 )= 25* 169

a2x2+a2y2+b2x2+b2y2= 25×169a^2x^2+a^2y^2+b^2x^2+b^2y^2=\ 25\times 169

k2= 652  (25× 169)k^2=\ 65^2\ -\ \left(25\times\ 169\right)

k = 0

D is the correct answer.

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