Find the 15

Find the tangent line of the ellipse 9 x² + 16 y² = 144 that has the slope k = -1

Result

t₁ = (Correct answer is: x + sqrt(19))
t₂ = (Correct answer is: x-sqrt(19))

Solution:

$$\begin{gathered} 9x^2+16y^2=144 \\ (x\mathrm{/}a{)}^2+(y\mathrm{/}b{)}^2=1 \\ a=\sqrt{16}=4 \\ b=\sqrt{3}=\sqrt{3}\doteq 1.7321 \\ k=-1 \\ {b}^2-q^2+a^2\ast k^2=0 \\ 1.73205080757^2-q^2+4^2\cdot (-1{)}^2=0 \\ -q^2+19=0 \\ {q}^2-19=0 \\ a=1;b=-0;c=-19 \\ D=b^2-4ac=0^2-4\cdot 1\cdot (-19)=76 \\ D>0 \\ {q}_{1,2}={\displaystyle \frac{-b\pm \sqrt{D}}{2a}}={\displaystyle \frac{\pm \sqrt{76}}{2}}=\pm \sqrt{19} \\ {q}_{1,2}=\pm \sqrt{19}=\pm 4.35889894354 \\ {q}_1=\sqrt{19}=4.35889894354 \\ {q}_2=-\sqrt{19}=-4.35889894354 \\ \text{ Factored form of the equation: } \\ (q-4.35889894354)(q+4.35889894354)=0 \\ t:y=k_x+q \\ {t}_1=x+\sqrt{19} \end{gathered}$$

$t_2=x-\sqrt{19}$

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