conics question (1 Viewer)

[cd285]

find the equation of two tangents to the ellipse 16X^2 + 25Y^2 =400 which are parallel to the line y= x+ 2

cheers
cd

 

[MC Squidge]

^
show working m8

 

[Timothy.Siu]

just find dy/dx and u know the gradient of the tangent u want to find is 1
so u make dy/dx=1

 

[LordPc]

derive this
16X^2 + 25Y^2 =400
solve for dy/dx = 1 (gradient of y=x+2)

 

[cd285]

i did that and you get -16x/25y=1

making y the subject u get -16x/25

then u would sub it into the original eqn...

but then it doesn't work


the answer is

x + or - the sq.root of 41

 

[azureus88]

You got x^2/25 + y^2/16 =1

Sub in y=mx+b and you get a quadratic in x. Make the discriminant equal to 0. You now have the general tangent. It should be something like y=mx+-sqrt(16m^2 + 25).

Since m=1, y=x+-sqrt(41)

 

[jet]

You got x^2/25 + y^2/16 =1

Sub in y=mx+b and you get a quadratic in x. Make the discriminant equal to 0. You now have the general tangent. It should be something like y=mx+-sqrt(16m^2 + 25).

Since m=1, y=x+-sqrt(41)

Exactly. Ill do the soln now.

 

[jet]

 

[jet]

and, in the interests of visualisation:

 

[gurmies]

and, in the interests of visualisation:

You love it xD

 

[GUSSSSSSSSSSSSS]

lol too true !! xD

 

[cd285]

thanks guys, much appreciated.
cd