C1 and C2 are two non-intersecting curves. Common normal of C1 and C2 intersect at the points (1,0) and (4, 4) on the curves. If the slope of the normal is 30 degrees, find the shortest distance between the curves assuming they have only one common normal

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$5 tan (30 °)$

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$5 sin (30 °)$

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$5 cos (30 °)$

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Solution

The correct option is A
5

Shortest distance between two curves is the length of the normal between them. It is also the distance between the points where the normal cuts the curves. We are given that those two points are (1,0) and (4,4).

Distance between them or shortest between the curves =

$\sqrt{(4 - 1)^{2} + (4 - 0)^{2}} = 5$

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C1 and C2 are two non-intersecting curves. Common normal of C1 and C2 intersect at the points (1,0) and (4, 4) on the curves. If the slope of the normal is 30 degrees, find the shortest distance between the curves assuming they have only one common normal

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$C_{1} : z + ¯ z = 2 | z - 1 |$
$C_{2} : a r g (z + 1 + i) = α$
Where $α$ belongs to the interval $(0, \frac{π}{2})$ such that curves $C_{1}$ and $C_{2}$ have exactly one point in common and which is denoted by $P (z_{0})$
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