Find the point on the curve x2 =8y which is nearest to the point (2,4) - Maths -

Question

Find the point on the curve x2 =8y which is nearest
to the point (2,4)

Santosh Kumar · Accepted Answer

Let (h, k) be any point on the curve x2 = 8y so, (h, k) satisfies the equation x2 = 8y $h^{2} = 8 k \Rightarrow k = \frac{h^{2}}{8}$ Now, distance between (h, k) and (2, 4), $D = \sqrt{(h - 2)^{2} + (k - 4)^{2}}$ (Distance formula between two points) $\Rightarrow D = \sqrt{(h - 2)^{2} + (\frac{h^{2}}{8} - 4)^{2}} \Rightarrow D^{2} = {(h - 2)^{2} + (\frac{h^{2}}{8} - 4)}^{2} \Rightarrow 2 D (\frac{d D}{d h}) = 2 (h - 2) + 2 (\frac{h^{2}}{8} - 4) (\frac{2 h}{8}) \Rightarrow (\frac{d D}{d h}) = \frac{2}{2 D} (\frac{h^{2}}{8} - 4) (\frac{h}{4})$ For the required point to be nearest to the point (2, 4), distance D should be minimum $\frac{d D}{d h} = 0$ $\Rightarrow \frac{2}{2 D} (\frac{h^{2}}{8} - 4) (\frac{h}{y}) = 0 \Rightarrow (\frac{h^{2} - 32}{8}) (\frac{h}{y}) = 0 \Rightarrow h = 0 or h^{2} = 32 \Rightarrow h = 0 or h = 4 \sqrt{2} k = \frac{h^{2}}{8} = \frac{32}{8} = 4$ âˆ´ Required point = $(h, k) = (4 \sqrt{2}, 4)$

Find the point on the curve x2 =8y which is nearest to the point (2,4).

Find the point on the curve x² =8y which is nearest to the point (2,4).