Hyperbola
Hello students, in this blog we are going to learn how to draw hyperbola in very easy step. According to google, Hyperbola is a symmetrical open curve formed by the intersection of a circular cone with a plane at a smaller angle with its axis than the side of the cone. Lets see how can we draw it.
Question: Point F is 50 mm from a line AB. a point P is moving in a plane such that the ratio of it's distances from F and line AB remains constant and equal to 2/3. draw locus of point P. { ECCENTRICITY = 2/3 }.Solution:
- Draw a vertical line AB and point F 50 mm from it.
- Divide 50 mm distance in 5 parts.
- Name 2nd part from F as V. It is 20mm and 30mm from F and AB line respectively. It is first point giving ratio of it’s distances from F and AB 2/3 (i.e 20/30)
- Form more points giving same ratio such as 30/45, 40/60, 50/75 etc.
- Taking 45,60 and 75mm distances from line AB, draw three vertical lines to the right side of it.
- Now with 30, 40 and 50mm distances in compass cut these lines above and below, with F as center.
- Join these points through V in smooth curve.
- Hyperbola is ready.
Practice Questions..
- Point F is 40 mm from a line AB. a point P is moving in a plane such that the ratio of it's distances from F and line AB remains constant and equal to 2/3. draw locus of point P. { ECCENTRICITY = 2/3 }.
- Point F is 60 mm from a line AB. a point P is moving in a plane such that the ratio of it's distances from F and line AB remains constant and equal to 2/3. draw locus of point P. { ECCENTRICITY = 2/3 }.
In closing..
I hope you understand the question. You can draw Hyperbola by simply applying the similar method. If you have any doubts then just let me know in the gmail! If you have any Chemistry related problem, see our latest blogs on it!