A parabola with equation y = (x^2 /k) − 5 intersects a circle with equation x^2 + y^2 = 25 at exactly 3 points A, B, C Determine all such positive integers k for which the area of ΔABC is an integer


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Question Number 11607 by Joel576 last updated on 29/Mar/17

$A parabola with equation y = \frac{x^{2}}{k} - 5 intersects$ $a circle with equation x^{2} + y^{2} = 25 at exactly 3 points A, B, C$ $Determine all such positive integers k for which$ $the area of Δ A B C is an integer$
	Answered by mrW1 last updated on 29/Mar/17
	$A(−t,s) B(0,−5) C(t,s) ΔABC=A=((2t(s+5))/2)=t(s+5) t^2 +s^2 =25 s=(t^2 /k)−5 ks=t^2 −5k ks=25−s^2 −5k s^2 +ks+5(k−5)=0 s=((−k±(√(k^2 −4×5(k−5))))/2)=((−k±(k−10))/2)=−5,5−k s=−5⇒Point B s=5−k⇒t^2 =25−(5−k)^2 =k(10−k) t=(√(k(10−k))) (k<10) A=t(s+5)=(√(k(10−k)))(10−k)=n (n=integer{ k(10−k)(10−k)^2 =n^2 k(10−k)=((n/(10−k)))^2 =i^2 (i=integer) k^2 −10k+i^2 =0 k=((10±(√(100−4i^2 )))/2)=5±(√(25−i^2 )) i=0⇒k=0,10 (not ok, since k≠0,k<10) i=1⇒k=5±(√(24)) (no integer) i=2⇒k=5±(√(21)) (no integer) i=3⇒k=5±4=1,9 i=4⇒k=5±3=2,8 i=5⇒k=5 ⇒all possible positive integer values for k are 1,2,5,8,9.$
	$A (- t, s)$ $B (0, - 5)$ $C (t, s)$ $Δ A B C = A = \frac{2 t (s + 5)}{2} = t (s + 5)$ $t^{2} + s^{2} = 25$ $s = \frac{t^{2}}{k} - 5$ $k s = t^{2} - 5 k$ $k s = 25 - s^{2} - 5 k$ $s^{2} + k s + 5 (k - 5) = 0$ $s = \frac{- k \pm \sqrt{k^{2} - 4 \times 5 (k - 5)}}{2} = \frac{- k \pm (k - 10)}{2} = - 5, 5 - k$ $s = - 5 \Rightarrow P o i n t B$ $s = 5 - k \Rightarrow t^{2} = 25 - {(5 - k)}^{2} = k (10 - k)$ $t = \sqrt{k (10 - k)} (k < 10)$ $A = t (s + 5) = \sqrt{k (10 - k)} (10 - k) = n (n = i n t e g e r {$ $k (10 - k) {(10 - k)}^{2} = n^{2}$ $k (10 - k) = {(\frac{n}{10 - k})}^{2} = i^{2} (i = i n t e g e r)$ $k^{2} - 10 k + i^{2} = 0$ $k = \frac{10 \pm \sqrt{100 - 4 i^{2}}}{2} = 5 \pm \sqrt{25 - i^{2}}$ $i = 0 \Rightarrow k = 0, 10 (n o t o k, s i n c e k \neq 0, k < 10)$ $i = 1 \Rightarrow k = 5 \pm \sqrt{24} (n o i n t e g e r)$ $i = 2 \Rightarrow k = 5 \pm \sqrt{21} (n o i n t e g e r)$ $i = 3 \Rightarrow k = 5 \pm 4 = 1, 9$ $i = 4 \Rightarrow k = 5 \pm 3 = 2, 8$ $i = 5 \Rightarrow k = 5$ $\Rightarrow a l l p o s s i b l e p o s i t i v e i n t e g e r v a l u e s f o r k$ $a r e 1, 2, 5, 8, 9 .$
	Commented by Joel576 last updated on 29/Mar/17

	$t h a n k y o u v e r y m u c h$
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