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junior college 2 | H2 Maths
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differentiation and application
Date Posted: 3 years ago
Views: 353
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done
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Find the first derivative of the equation to find the gradient of the line, equate to the one given and solve for teeta, and subsequently the x and y coordinates.
Date Posted:
3 years ago
done
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Find the equation of the tangent to find the x and y intercepts of the line.
With the coordinates, find the coordinates of the midpoint
Derive a parametric equation for the mid point and solve for the cartesian equation.
(Hihi, not too sure if there will be any careless mistake so do take note.)
With the coordinates, find the coordinates of the midpoint
Derive a parametric equation for the mid point and solve for the cartesian equation.
(Hihi, not too sure if there will be any careless mistake so do take note.)
Date Posted:
3 years ago
done
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Find the equation of the tangent to find the x and y intercepts of the line.
With the coordinates, find the coordinates of the midpoint
Derive a parametric equation for the mid point and solve for the cartesian equation.
Check with GC.
(Hihi, not too sure if there will be any careless mistake so do take note.)
With the coordinates, find the coordinates of the midpoint
Derive a parametric equation for the mid point and solve for the cartesian equation.
Check with GC.
(Hihi, not too sure if there will be any careless mistake so do take note.)
Date Posted:
3 years ago
hi! I am able to solve till the mid point. But I don't know what to do with the mid point. In your answer (last picture), where did the y² come from and why y²?
hey! so basically the mid point coordinates will be following the two parametric equations as seen in the picture, the x and y coordinates.
the y^2 is just a method to fit them into a cartesian equation
the y^2 is just a method to fit them into a cartesian equation
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