• Best of luck to the class of 2024 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here
  • YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page
MedVision ad

Conics Question Help Please! (1 Viewer)

a1079atw

Member
Joined
Mar 2, 2013
Messages
60
Gender
Undisclosed
HSC
2014
Find the equation of the tangents drawn from the point (6, –2) to the ellipse 4x2 + 9y2 = 36.

Thanks!
 

esaitchkay

Active Member
Joined
Jun 12, 2012
Messages
220
Gender
Male
HSC
2014
you can either:
i) use the equation of tangent for an ellipse
and then just sub in (6,-2) as x1 and y1, a=3 and b=2

or if that's unknown (or you don't know how to find it yet)
ii) implicitly differentiate the equation ( http://www.sosmath.com/calculus/diff/der05/der05.html ) - dy/dx will give you the gradient; sub the point (6,-2) in.
then just use point-gradient formula: y-y1=dy/dx(x-x1)

ii) is the method you use to find the result in i)

oops I read the question wrong. LOL
I suppose you can just find the chord of contact where (x0,y0) is (6,-2) and then find the point of intersection (make y the subject and sub the equation into the ellipse) and sub that into the tangent to ellipse equation. Here's what I came up with: https://www.dropbox.com/s/5wfhk56i9klvmx0/Scan0004.jpg but hopefully someone has a shorter and smarter method haha.
 
Last edited:

dunjaaa

Active Member
Joined
Oct 10, 2012
Messages
473
Gender
Male
HSC
2014
Just a tip, with a sketch, you would of been able to determine one of the tangents straight away (y=-2)
 

a1079atw

Member
Joined
Mar 2, 2013
Messages
60
Gender
Undisclosed
HSC
2014
you can either:
i) use the equation of tangent for an ellipse
and then just sub in (6,-2) as x1 and y1, a=3 and b=2

or if that's unknown (or you don't know how to find it yet)
ii) implicitly differentiate the equation ( http://www.sosmath.com/calculus/diff/der05/der05.html ) - dy/dx will give you the gradient; sub the point (6,-2) in.
then just use point-gradient formula: y-y1=dy/dx(x-x1)

ii) is the method you use to find the result in i)

oops I read the question wrong. LOL
I suppose you can just find the chord of contact where (x0,y0) is (6,-2) and then find the point of intersection (make y the subject and sub the equation into the ellipse) and sub that into the tangent to ellipse equation. Here's what I came up with: https://www.dropbox.com/s/5wfhk56i9klvmx0/Scan0004.jpg but hopefully someone has a shorter and smarter method haha.
Thank you! I like your way of setting out the solution, clear and easy to follow.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top