If the point on the curve y^{2}=6x,nearest to | vedclass

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(B) The minimum distance from a point to a curve is along the normal to the curve at that point.Let the point on the parabola $y^{2}=6x$ be $P\left(\f

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(B) The minimum distance from a point to a curve is along the normal to the curve at that point.Let the point on the parabola $y^{2}=6x$ be $P\left(\frac{3}{2}t^{2}, 3t\right)$,where $4a=6 \Rightarrow a=\frac{3}{2}$.The equation of the normal at point $P(t)$ is $tx + y = 2at + at^{3}$.Substituting $a=\frac{3}{2}$,the normal equation is $tx + y = 3t + \frac{3}{2}t^{3}$.Since this normal passes through the point $\left(3, \frac{3}{2}\right)$,we have:$t(3) + \frac{3}{2} = 3t + \frac{3}{2}t^{3}$$3t + \frac{3}{2} = 3t + \frac{3}{2}t^{3}$$\frac{3}{2} = \frac{3}{2}t^{3}$$t^{3} = 1 \Rightarrow t = 1$.Thus,the point $P$ is $\left(\frac{3}{2}(1)^{2}, 3(1)\right) = \left(\frac{3}{2}, 3\right)$.So,$\alpha = \frac{3}{2}$ and $\beta = 3$.The value of $2(\alpha+\beta) = 2\left(\frac{3}{2} + 3\right) = 2\left(\frac{9}{2}\right) = 9$.

If the point on the curve $y^{2}=6x$,nearest to the point $\left(3, \frac{3}{2}\right)$ is $(\alpha, \beta)$,then $2(\alpha+\beta)$ is equal to $.....$

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