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Question Number 32508 by Tinkutara last updated on 26/Mar/18

Answered by MJS last updated on 27/Mar/18

$$\mathrm{there}\mathrm{are}2\mathrm{tangents}\mathrm{of}\mathrm{different}$$$$\mathrm{lengths}\mathrm{but}\mathrm{none}\mathrm{of}\mathrm{these}\mathrm{answers}$$$$\mathrm{fit}\mathrm{any}\mathrm{of}\mathrm{them}...$$$$\mathrm{ellipse}:y=\pm \frac{b}{a}\sqrt{a^2-x^2}$$$$\mathrm{tangent}\mathrm{in}\mathrm{point}\left(\begin{array}{c}o\\ \pm \frac{b}{a}\sqrt{a^2-o^2}\end{array}\right):$$$$t:y=\mp \frac{bo}{a\sqrt{a^2-o^2}}\times x\pm \frac{ab}{\sqrt{a^2-o^2}}$$$$\mathrm{tangents}\mathrm{through}\mathrm{given}\mathrm{point}$$$$P=\left(\begin{array}{c}p\\ q\end{array}\right):\mathrm{solve}\mathrm{above}\mathrm{equations}$$$$\mathrm{for}o\mathrm{with}x=p\wedge y=q$$$$\mathrm{in}\mathrm{this}\mathrm{case}:$$$$a=\sqrt{3};b=2;p=3;q=-5$$$$-5=\mp \frac{2\mathrm{o}}{\sqrt{3}\sqrt{3-o^2}}\times 3\pm \frac{2\sqrt{3}}{\sqrt{3-o^2}}$$$$o_{1(y<0)}=\frac{12}{37}-\frac{15\sqrt{11}}{37}$$$$o_{2(y>0)}=\frac{12}{37}+\frac{15\sqrt{11}}{37}$$$$\mathrm{the}\mathrm{tangents}:$$$$t_1:y=-(\frac{5}{2}-\frac{\sqrt{11}}{2})\times x+(\frac{5}{2}-\frac{3\sqrt{11}}{2})$$$$t_2:y=-(\frac{5}{2}+\frac{\sqrt{11}}{2})\times x+(\frac{5}{2}+\frac{3\sqrt{11}}{2})$$$$\mathrm{the}2\mathrm{points}\mathrm{of}\mathrm{the}\mathrm{ellipse}\mathrm{are}:$$$$O_1=\left(\begin{array}{c}\frac{12}{37}-\frac{15\sqrt{11}}{37}\\ -\frac{20}{37}-\frac{12\sqrt{11}}{37}\end{array}\right)$$$$O_2=\left(\begin{array}{c}\frac{12}{37}+\frac{15\sqrt{11}}{37}\\ -\frac{20}{37}+\frac{12\sqrt{11}}{37}\end{array}\right)$$$$\mathrm{distances}\mathrm{to}P=\left(\begin{array}{c}3\\ -5\end{array}\right)$$$$\mid P-{\mathrm{O}}_1\mid =\frac{3}{37}\sqrt{55(83-2\sqrt{11})}\approx 5.25$$$$\mid P-{\mathrm{O}}_2\mid =\frac{3}{37}\sqrt{55(83+2\sqrt{11})}\approx 5.69$$$$\mathrm{the}\mathrm{options}\mathrm{are}:$$$$\frac{\sqrt{17}}{2}\approx 2.06$$$$\frac{\sqrt{33}}{2}\approx 2.87$$$$\frac{\sqrt{37}}{2}\approx 3.04$$$$\frac{\sqrt{19}}{2}\approx 2.18$$

Commented by Tinkutara last updated on 27/Mar/18

Commented by Tinkutara last updated on 27/Mar/18

@MJS you are right. Graphing calculator confirms your answer. Thank you Sir! ��������

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