Draw a hyperbola when the distance of its focus from directrix is 65 mm and eccentricity is 4/3. Draw a tangent and normal to the curve at a distance of 35 mm from directrix. Try this one
As of now I'm sitting in the engineering drawing lab at IIT kharagpur watching this cuz the TA's don't give a fuck MMM hall just in case another kgpian is watching
Draw a conic curve whose distance of focus from directrix is 55 mm and eccentricity=1.5.Also draw tangent and normal at a distance of 50 mm from the fixed line
It's not meant to make an ellipse. The eccentricity of a hyperbola is 3/2 ( distance from focus divided by perpendicular distance from directrix),so that's why the vertex started at the second division. If it had started from the 3rd division, it would have meant the eccentricity was 2/3, and then it would have produced an ellipse
Very ausome direct teaching no bakwaas👍
Very Useful For Last Minute Revisions 👌
awesome explnation sir thank uh so much sir 😇😇
Draw a hyperbola when the distance of its focus from directrix is 65 mm
and eccentricity is 4/3. Draw a tangent and normal to the curve at a distance
of 35 mm from directrix.
Try this one
Construct Hyperbola Curve When Distance of Focus from Directrix is 45 mm and Eccentricity 2/1.
Exactly!! Everyone is using 50mm. Pls use another dimension like 40 or 60mm
As of now I'm sitting in the engineering drawing lab at IIT kharagpur watching this cuz the TA's don't give a fuck
MMM hall just in case another kgpian is watching
Thanks very much…please what app are you using to teach?
Abe parabola aur hyperbola ek hi jaisa tune nikala hai😂
Please make a video to draw hyperbola by vertices focus method
Vertices and focus are given
@Abdul Majid Ali Muhammad. Thank You for your feedback. We will let our backend team know.
Draw a conic curve whose distance of focus from directrix is 55 mm and eccentricity=1.5.Also draw tangent and normal at a distance of 50 mm from the fixed line
.Draw a conic curve when the distance of the focus from the vertex is 45 mm and
the eccentricity e = 3/2. Name the curv
Hyperbola
Thanku very much sir
Draw a hyperbola when the distance its focus and directrix is
50mm and eccentricity is 3/2.
Very good
What is answer it is 60mm
Aousam sir from
Gujrat...
Thank you so much sir
Nyc
Kul
Why divide CF in 5 equal parts?
Because of 3+ 2 : 5
Ve distance
wrong video this will make an ellipse not a hyperbola because V is marked wrong
It's not meant to make an ellipse. The eccentricity of a hyperbola is 3/2 ( distance from focus divided by perpendicular distance from directrix),so that's why the vertex started at the second division. If it had started from the 3rd division, it would have meant the eccentricity was 2/3, and then it would have produced an ellipse
😂 you are not drawing correctly
VF should be 3 not 2 units.
@What_on_earth_is_this
Remember it is three divisions from the focus not from C. That makes what you are saying wrong.
Thank you so much
Thank you so much Sir🙏