Question:
If the point on the curve $\mathrm{y}^{2}=6 \mathrm{x}$, nearest to the point $\left(3, \frac{3}{2}\right)$ is $(\alpha, \beta)$, then $2(\alpha+\beta)$ is equal to
Solution:
$\mathrm{P} \equiv\left(\frac{3}{2} \mathrm{t}^{2}, 3 \mathrm{t}\right)$
Normal at point $P$
$\mathrm{tx}+\mathrm{y}=3 \mathrm{t}+\frac{3}{2} \mathrm{t}^{3}$
Passes through $\left(3, \frac{3}{2}\right)$
$\Rightarrow 3 \mathrm{t}+\frac{3}{2}=3 \mathrm{t}+\frac{3}{2} \mathrm{t}^{3}$
$\mathrm{P} \equiv\left(\frac{3}{2}, 3\right)=(\alpha, \beta)$
$\Rightarrow \mathrm{t}^{3}=1 \Rightarrow \mathrm{t}=1$
$2(\alpha+\beta)=2\left(\frac{3}{2}+3\right)=9$
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