Tangents to ellipse

Find the magnitude of the angle at which the ellipse x² + 5 y² = 5 is visible from the point P[5, 1] .

Correct result:

A =  26.5651 °

Solution:

$$\begin{gathered} x^2+5y^2=5 \\ {P}_0=5;P_1=1 \\ {x}^2\mathrm{/}5+(y\mathrm{/}1{)}^2=1 \\ a=\sqrt{5}=\sqrt{5}\doteq 2.2361 \\ b=1 \\ {x}_0=0 \\ {y}_0=0 \\ {p}_1,p_2:y=k_x+q \\ {b}^2-q^2+a^2\cdot k^2=0 \\ 1-q^2+5\cdot k^2=0 \\ {P}_1=k\cdot P_0+q \\ 1-(P_1-k\ast P_0{)}^2+5\ast k^2=0 \\ 1-(1-k\cdot 5{)}^2+5\cdot k^2=0 \\ -20k^2+10k=0 \\ 20k^2-10k=0 \\ a=20;b=-10;c=0 \\ D=b^2-4ac=10^2-4\cdot 20\cdot 0=100 \\ D>0 \\ {k}_{1,2}={\displaystyle \frac{-b\pm \sqrt{D}}{2a}}={\displaystyle \frac{10\pm \sqrt{100}}{40}} \\ {k}_{1,2}={\displaystyle \frac{10\pm 10}{40}} \\ {k}_{1,2}=0.25\pm 0.25 \\ {k}_1=0.5 \\ {k}_2=0 \\ \text{ Factored form of the equation: } \\ 20(k-0.5)k=0 \\ {q}_1=P_1-k_1\cdot P_0=1-0.5\cdot 5=-{\displaystyle \frac{3}{2}}=-1.5 \\ {q}_2=P_1-k_2\cdot P_0=1-0\cdot 5=1 \\ {\phi }_1={\displaystyle \frac{180^{\circ }}{\pi }}\cdot \arctan (k_1)={\displaystyle \frac{180^{\circ }}{\pi }}\cdot \arctan (0.5)\doteq 26.5651^{\circ } \\ {\phi }_2={\displaystyle \frac{180^{\circ }}{\pi }}\cdot \arctan (k_2)={\displaystyle \frac{180^{\circ }}{\pi }}\cdot \arctan (0)=0^{\circ } \\ A=\mathrm{\mid }{\phi }_1-{\phi }_2\mathrm{\mid }=\mathrm{\mid }26.5651-0\mathrm{\mid }=26.5651^{\circ }=26^{\circ }33^{\mathrm{\prime }}54\mathrm{"} \end{gathered}$$

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