Question

the distance between the straight lines xcosalpha+ysinalpha=sec^alpha and xcosalpha+ysinalpha=tan^alpha is .​

Answer 1

Answer:

......,..............................................

Answer 2

Answer: The distance between the lines is 1 unit.

Step-by-step explanation:

L1  ≡  xcosα + ysinα = sec²α

L2 ≡  xcosα + ysinα = tan²α

Both the lines have the same slope -cotα, hence the given lines are parallel.

Distance between parallel lines is always fixed and they never meet each other.

Let there be two parallel lines be of form ax + by =  c1 and ax + by = c2

The distance d between the parallel lines is given by the formula :

$d = \frac{|c1 - c2|}{\sqrt{a^{2} +b^{2} } }$

For L1 and L2 :

a = cosα

b = sinα

c1 = sec²α

c2 = tan²α

Distance d between L1 and L2 is given by :

$d = \frac{|sec^{2}\alpha - tan^{2}\alpha |}{\sqrt{cos^{2}\alpha +sin^{2}\alpha } }$

$d = \frac{|1|}{\sqrt{1 } }$

d = 1 units

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