hard ellipse qn? (1 Viewer)

[blakwidow]

Show that if y = mx+k is a tangent to the ellipse at

+

= 1, then

soln.
The ellipse has parametric eqn

and

.

Hence

=

. If y =mx+k at

, then

m=

at P

ma

+

= 0
.
.
.
how did they obtain the last line?
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=150052

 

[Ea22.007]

i dont think thats the best way to solve this one.
you should maybe try subbing in y=mx+k in the equation of the ellipse for y and then rearrange to get a quadratic in terms of f(x)=0

then take the discriminant of that quadratic and let it =0 as it is a tangent after that just play around with the algebra and you can rearrange to get the required statement.

 

[victorheaven]

y=mx+k. m is the gradient of the tangent.
but dy/dx=-b*cos/a*sin and dy/dx is the gradient too
so m=-b*cos/a*sin
m*a*sin=-b*cos
m*a*sin+b*coz=0