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-ABC-is-an-integer

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

$$\mathrm{A}\mathrm{parabola}\mathrm{with}\mathrm{equation}y=\frac{x^2}{k}-5\mathrm{intersects}$$$$\mathrm{a}\mathrm{circle}\mathrm{with}\mathrm{equation}{x}^2+y^2=25\mathrm{at}\mathrm{exactly}3\mathrm{points}A,B,C$$$$\mathrm{Determine}\mathrm{all}\mathrm{such}\mathrm{positive}\mathrm{integers}k\mathrm{for}\mathrm{which}$$$$\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{\Delta }ABC\mathrm{is}\mathrm{an}\mathrm{integer}$$

Answered by mrW1 last updated on 29/Mar/17

$$A(-t,s)$$$$B(0,-5)$$$$C(t,s)$$$$\mathrm{\Delta }ABC=A=\frac{2t(s+5)}{2}=t(s+5)$$$$t^2+s^2=25$$$$s=\frac{t^2}{k}-5$$$$ks=t^2-5k$$$$ks=25-s^2-5k$$$$s^2+ks+5(k-5)=0$$$$s=\frac{-k\pm \sqrt{k^2-4\times 5(k-5)}}{2}=\frac{-k\pm (k-10)}{2}=-5,5-k$$$$s=-5\Rightarrow PointB$$$$$$$$s=5-k\Rightarrow t^2=25-{(5-k)}^2=k(10-k)$$$$t=\sqrt{k(10-k)}(k<10)$$$$A=t(s+5)=\sqrt{k(10-k)}(10-k)=n(n=integer\{$$$$k(10-k){(10-k)}^2=n^2$$$$k(10-k)={\left(\frac{n}{10-k}\right)}^2=i^2(i=integer)$$$$k^2-10k+i^2=0$$$$k=\frac{10\pm \sqrt{100-4i^2}}{2}=5\pm \sqrt{25-i^2}$$$$i=0\Rightarrow k=0,10(notok,sincek\ne 0,k<10)$$$$i=1\Rightarrow k=5\pm \sqrt{24}\left(nointeger\right)$$$$i=2\Rightarrow k=5\pm \sqrt{21}\left(nointeger\right)$$$$i=3\Rightarrow k=5\pm 4=1,9$$$$i=4\Rightarrow k=5\pm 3=2,8$$$$i=5\Rightarrow k=5$$$$$$$$\Rightarrow allpossiblepositiveintegervaluesfork$$$$are1,2,5,8,9.$$

Commented by Joel576 last updated on 29/Mar/17

$$thankyouverymuch$$

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