Conics Query (2 Viewers)

[MC Squidge]

P(acos@,bsin@) lies on the ellipse (x^2)/(a^2)+(y^2)/(b^2)=1. The Normal at P cuts the x axis at X and the Y axis at Y. Show that (PX)/(PY)=(b^2)/(a^2)

is the distance formula the only way to go about this one?

 

[shaon0]

P(acos@,bsin@) lies on the ellipse (x^2)/(a^2)+(y^2)/(b^2)=1. The Normal at P cuts the x axis at X and the Y axis at Y. Show that (PX)/(PY)=(b^2)/(a^2)

is the distance formula the only way to go about this one?

yeah, i think thats the way to go.

 

[MC Squidge]

i was told to use the distance formula as a last resort, as there is usually an easier method

 

[gurmies]

Seems right for this question. Avoid using the distance formula where utilising the simple "focus directrix" definition will suffice. In this case there is no mention of either foci or directricies, so distance formula all the way i'd say.

 

[shaon0]

i was told to use the distance formula as a last resort, as there is usually an easier method

there probably is but it may take sometime to figure it out.

 

[seanieg89]

Solution without Distance Formula:

Let be the projection of onto the y-axis.

ie

It is fairly simple to show that is similar to



Note and are simply the x-coordinates of and . And finding the point is routine.

This is valid as long as . The argument can be tailored to suit the alternate case.


Similar triangles can make some conic calculations a whole lot simpler.

 

[undalay]

RP is the x coordinate of point P ( acos@)
RX is the x axis intercept of the normal line.

TP is RP - RX

TPX is similar to RPY

PX/PY = TP/RP

For most ratio prove questions, you can incorporate similar triangles into the question. Try it next time : )

edit: Sometimes u can get away without proving the similarity of triangles if u use the intercept theorem (and quote it) . http://en.wikipedia.org/wiki/Intercept_theorem

edit edit: Guy above beat me to it, gw: )

 

[MC Squidge]

thanks

 

[Timothy.Siu]

sub in y=mx+c into the equation and then if its a tangent theres one root, so discriminant=0 and you'll get it

 

[Timothy.Siu]

damn, you know i tired that and i still got it wrong =/

i sorta got it,
u get to,

x²b²+m²x²a²+2mxca²+a²c²-a²b²=0

x²(b²+a²m²)+2mca²x+a²c²-a²b²=0

for one root, discriminant=0

discriminant= 4m²c²a⁴-4(b²+a²m²)(a²c²-a²b²)

=4a²(m²c²a²-(b²+a²m²)(c²-b²) =0

m²c²a²-(b²c²-b⁴+a²m²c²-a²m²b²)=0

then u get c²b²=a²m²b²+b²⁴=0
c²=a²m²+b²

 

[GUSSSSSSSSSSSSS]

b^4 not b^2 in the second last line right?

lol it actually says b^24

but yes b^4 is correct not b^2

 

[Timothy.Siu]

lol, yeah.

 

[untouchablecuz]

Mind if I ask a question on here to aviod making a new board?

Show that the line y = mx + c is a tangent to the ellipse x^2/a^2 + y^2/b^2 = 1 if c^2 = a^2m^2 + b^2

Thanks.

do it using the parametric equations of an ellipse, much easier than the cartesian method

x=acos0

dx/d0 = -asin0

y = bsin0

dy/d0 = bcos0

.'. dy/dx = (-b/a)(cos0/sin0)

If the line y = mx +b is tangent:

m = (-b/a)(cos0/sin0).............(1)

also, y = bsin0 and x=acos0 satisfy y = mx +b

hence,

bsin0 = macos0 + b.............(2)

solve (1) and (2) simulataneously eliminating 0 (theta), and you'll get the expression