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$a x-y+c=0$ is the equation of the common tangent to the parabola $y^2=8 \sqrt{5} x$ and the circle $x^2+y^2=1$. If this tangent makes an acute angle with the positive $\mathrm{X}$ - axis in the positive direction, then $\mathrm{a}^2 \mathrm{c}^2=$
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Verified Answer
The correct answer is:
20
$T: y=a x+C$
for $y^2=8 \sqrt{5} x \Rightarrow C=\frac{2 \sqrt{5}}{a}$
for $x^2+y^2=1 \Rightarrow C= \pm \sqrt{1+a^2}$
$\Rightarrow C^2=1+a^2=\left(\frac{2 \sqrt{5}}{a}\right)^2 \Rightarrow 1+a^2=\frac{20}{a^2}$ Let $a^2=t$
Let $a^2=t$
$$
\begin{aligned}
& \Rightarrow \quad t+1-\frac{20}{t}=0 \Rightarrow t^2+i-20=0 \\
& \Rightarrow \quad(t+5)(t-4)=0 \Rightarrow t=4,-5 \\
& \therefore \quad a^2=4 \Rightarrow c^2=1+a^2=5 \\
& \therefore \quad a^2 c^2=4.5=20 .
\end{aligned}
$$
for $y^2=8 \sqrt{5} x \Rightarrow C=\frac{2 \sqrt{5}}{a}$
for $x^2+y^2=1 \Rightarrow C= \pm \sqrt{1+a^2}$
$\Rightarrow C^2=1+a^2=\left(\frac{2 \sqrt{5}}{a}\right)^2 \Rightarrow 1+a^2=\frac{20}{a^2}$ Let $a^2=t$
Let $a^2=t$
$$
\begin{aligned}
& \Rightarrow \quad t+1-\frac{20}{t}=0 \Rightarrow t^2+i-20=0 \\
& \Rightarrow \quad(t+5)(t-4)=0 \Rightarrow t=4,-5 \\
& \therefore \quad a^2=4 \Rightarrow c^2=1+a^2=5 \\
& \therefore \quad a^2 c^2=4.5=20 .
\end{aligned}
$$
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