Suppose that the foci of the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{5}=1$ are $(f_{1},0)$ and $(f_{2},0)$ where $f_{1}>0$ and $f_{2}<0$. Let $P_{1}$ and $P_{2}$ be two parabolas with a common vertex at $(0,0)$ and with foci at $(f_{1},0)$ and $(2f_{2},0)$, respectively. Let $T_{1}$ be a tangent to $P_{1}$ which passes through $(2f_{2},0)$ and $T_{2}$ be a tangent to $P_{2}$ which passes through $(f_{1},0)$. If $m_{1}$ is the slope of $T_{1}$ and $m_{2}$ is the slope of $T_{2}$, then the value of $\left(\frac{1}{m_{1}^{2}}+m_{2}^{2}\right)$ is

Suppose that the foci of the ellipse x²/9+y²/5=1 are (f₁,0) and (f₂,0) where f₁>0 and f₂<0. Let P₁ and P₂ be two parabolas with a common vertex at (0,0) and with foci at (f₁,0) and (2f₂,0), respectively. Let T₁ be a tangent to P₁ which passes through (2f₂,0) and T₂ be a tangent to P₂ which passes through (f₁,0). If m₁ is the slope of T₁ and m₂ is the slope of T₂, then the value of (1/m₁²+m₂²) is $\text{Ellipse: } \dfrac{x^{2}}{9}+\dfrac{y^{2}}{5}=1$ $\text{foci } (f_{1},0), (f_{2},0); \quad f_{1}>0,\ f_{2}<0$ $P_{1}, P_{2}: \text{ vertex } (0,0), \text{ foci } (f_{1},0), (2f_{2},0)$ $T_{1}: \text{ tangent to } P_{1} \text{ through } (2f_{2},0), \quad \text{slope } m_{1}$ $T_{2}: \text{ tangent to } P_{2} \text{ through } (f_{1},0), \quad \text{slope } m_{2}$ $\text{Find: } \dfrac{1}{m_{1}^{2}}+m_{2}^{2}$ $\dfrac{x^{2}}{9}+\dfrac{y^{2}}{5}=1 \Rightarrow a=3,\ b=\sqrt{5}$ $c^{2}=a^{2}-b^{2}=9-5=4 \Rightarrow c=2$ $\text{foci} = (\pm c,0) = (\pm 2,0)$ $\therefore f_{1}=2, \quad f_{2}=-2$ $P_{1}: \text{vertex } (0,0), \text{ focus } (2,0) \Rightarrow a=2$ $y^{2}=4ax=8x$ $P_{2}: \text{focus } (2f_{2},0)=(-4,0) \Rightarrow |a|=4$ $y^{2}=-4|a|x=-16x$ $1.\ \text{tangent of } y^{2}=4ax:\ y=mx+\dfrac{a}{m}$ $2.\ T_{1} \text{ through } (-4,0) \Rightarrow m_{1}^{2}$ $3.\ T_{2} \text{ through } (2,0) \Rightarrow m_{2}^{2}$ $y^{2}=4ax \Rightarrow \text{tangent } y=mx+\dfrac{a}{m}$ $P_{1}: 4a=8 \Rightarrow a=2; \quad T_{1}: y=m_{1}x+\dfrac{2}{m_{1}}$ $(-4,0) \Rightarrow 0=-4m_{1}+\dfrac{2}{m_{1}}$ $4m_{1}^{2}=2 \Rightarrow m_{1}^{2}=\dfrac{1}{2}$ $\dfrac{1}{m_{1}^{2}}=2$ $T_{1}: y=\dfrac{1}{\sqrt{2}}x+2\sqrt{2}$ $P_{2}: 4a=-16 \Rightarrow a=-4; \quad T_{2}: y=m_{2}x-\dfrac{4}{m_{2}}$ $(2,0) \Rightarrow 0=2m_{2}-\dfrac{4}{m_{2}}$ $2m_{2}^{2}=4 \Rightarrow m_{2}^{2}=2$ $T_{2}: y=-\sqrt{2}x+2\sqrt{2}$ $\dfrac{1}{m_{1}^{2}}+m_{2}^{2}=2+2$ $\boxed{4}$