conics questions (1 Viewer)

[pLuvia]

Find the equation of the tangents to the ellipse 4x²+5y²=20 which are perpendicular to the line 2x+2y-1=0

Find the coordinates of the chords of contact of tangents from the point (1,-1) to the ellipse 5x²+2y²=3

 

[joesmith1975]

Start you off 4x squared + 5y squared = 20

/ by 20

x squared /5 + y squared / 4 =1

a squared =5 b squared = 4

a = root 5 b = root 4 = 2

Thats the ellispe.

The line 2x+2y-1=0
2y=-2x +1
y=-x +1/2

m of 1 = -1


therefore to fnd perpendicular to the line m of 1 times m of 2 = -1

m of 2 =1

using equation y=mx +and - root {a squared m squared + b squared}
y=x + and - root {5 times 1 + 4}
y= x + and - 3


i'll leave someone else to the next bit.

 

[Mountain.Dew]

using equation y=mx +and - root {a squared m squared + b squared}
y=x + and - root {5 times 1 + 4}
y= x + and - 3


i'll leave someone else to the next bit.

mmmmmm i dunno if what uve done is right...where did u get the " using equation y=mx +and - root {a squared m squared + b squared}" from? the a^2m^2 + b^2 DOES tickle my memory, what i want to know how u developed this 'general' equation, and for the benefit for all the BOS forum members here.

 

[pLuvia]

I don't understand

 

[joesmith1975]

you know how a line eg. y=3x+2 where m=3 and y-intecept = 2

its that type of formula {equation of tangent in terms of its graident}

The formula and how its devired it pretty long but ill do it, i dont think some teachers dont do it but my teacher is fully into maths and is a good teacher.


from equation of tangent using parametres {a cos θ , b sin θ}

x cos θ/a + y sin θ/b = 1

y sin θ = 1 - x cos θ/a

y= b/sin θ{1-x cos θ/a}

y= -bx cos θ/ sin θ + b / sin θ

y= -bx/a cot θ + b cosec θ

Therefore let m = -b/a cot θ

Now -am/b = cot θ

a squared m squared / b squared = cot squared θ

a squared m squared/ b squared = cosec squared θ -1 { from trig.}

cosec squared θ = 1 + a squared m squared / b squared

cosec squared θ = b squared + a squared m squared/ b squared

cosec θ = + and - root {b squared + a squared m squared / b squared}

As y=mx + b

y = mx +b + and - root {b squared + a squared m squared / b}

b's go away

therefore left with

y=mx + and - root {a squared m squared + b squared}



I know its long i hope you understand if not message back on post get back to you as soon as possible.

 

[_ShiFTy_]

For Q1, since the tangents are perpendicular to 2x + 2y =1 (y = -x + 1/2)
Then the tangents have a gradient of 1
So the family of lines that have a gradient of 1 is y = x + c

Sub y = x + c into 4x²+5y²=20
Muck around with it, and the discriminant delta must be equal to zero for it to be a tangent. You should get 2 values of c. Plug it back into the line and you will get ur equations..

 

[Rax]

For the 2nd part use the eqtn for chord of contact......I presume you now it

 

[Mountain.Dew]

comparing joesmith1975's and _ShiFTy_'s methods, i think that in this case, in the context of the question, _ShiFTy_'s is that little bit better. it is easier to follow, the logic seems more clearer AS you do the question. i base this opinion based upon the context of the question. joesmith1975's method works as well, but it would be far superior IF there was a previous part (of the same question) that said to show that the general equation of a tangent is:
y=mx + and - root {a squared m squared + b squared}.

but, this is only a personal response. it is really up 2 you to decide method you want to use.

 

[Riviet]

For the first one: what about implicitly differentiating the ellipse, letting dy/dx=1 to obtain -4x=5y, and substituting y=+sqrt[(20-4x²)/5] and solve for x to get x=+10/6, substitute x into ellipse to find the y co-ordinates, and use point gradient formula to find the equation of the tangents?

 

[pavittness]

a^2= 3/5 b^2= 3/2

using the result x.x0 /a^2 + y.y0/b^2 =1
sub in the point (1,-1) gives u

5x-2y=3

 

[Mountain.Dew]

For the first one: what about implicitly differentiating the ellipse, letting dy/dx=1 to obtain -4x=5y, and substituting y=+sqrt[(20-4x²)/5] and solve for x to get x=+10/6, substitute x into ellipse to find the y co-ordinates, and use point gradient formula to find the equation of the tangents?

mmmm that would work as well. however, just by mere observations, i think _ShiFTy_'s is quicker, in a sense.

also, riviet, when u said "...substitute x into ellipse to find the y co-ordinates, and use point gradient formula..." its much better to sub x=+10/6 into -4x=5y to get ur y-coordinate. then, as u said, use the point-gradient formula, knowing that m = 1 already

 

[Mountain.Dew]

a^2= 3/5 b^2= 3/2

using the result x.x0 /a^2 + y.y0/b^2 =1
sub in the point (1,-1) gives u

5x-2y=3

Pavittness, welcome to the BOS forums!

for the benefit of all the BOS forum members here, please show us the proof of getting
x.x0 /a^2 + y.y0/b^2 =1 as the equation of the chord of contact. im sure it will be helpful for many of us here.

 

[_ShiFTy_]

Pavittness, welcome to the BOS forums!

for the benefit of all the BOS forum members here, please show us the proof of getting
x.x0 /a^2 + y.y0/b^2 =1 as the equation of the chord of contact. im sure it will be helpful for many of us here.

Proving the chord of contact isnt that hard...
Say u have 2 tangents at P (x1, y1) and Q (x2, y2)
Tangent P = xx1/a^2 + yy1/b^2 = 1
Tangent Q = xx2/a^2 + yy2/b^2 = 1

From inspection, you can see that xx0/a^2 + yy0/b^2 = 1 satisfies both P and Q

 

[Riviet]

mmmm that would work as well. however, just by mere observations, i think _ShiFTy_'s is quicker, in a sense.

also, riviet, when u said "...substitute x into ellipse to find the y co-ordinates, and use point gradient formula..." its much better to sub x=+10/6 into -4x=5y to get ur y-coordinate. then, as u said, use the point-gradient formula, knowing that m = 1 already

Ah cool, my method is more 2u'ish, but isn't as messy I guess. Thanks for the tip.

 

[Mountain.Dew]

Ah cool, my method is more 2u'ish, but isn't as messy I guess. Thanks for the tip.

no probs. although i have to say, implicit diff isnt really 2u'ish at all...

Proving the chord of contact isnt that hard...
Say u have 2 tangents at P (x1, y1) and Q (x2, y2)
Tangent P = xx1/a^2 + yy1/b^2 = 1
Tangent Q = xx2/a^2 + yy2/b^2 = 1

From inspection, you can see that xx0/a^2 + yy0/b^2 = 1 satisfies both P and Q

ummmmm i dont think that's a sufficient enough proof...we are talking about proving something, not SHOWING something, where 'from inspection' would be a valid way.

but, i am not too sure myself. i prolly need to ask my maths teacher or do a little bit of research. i suppose its just that i have been taught the...'long' way...

Celebrating my 500th post!

 

[Riviet]

no probs. although i have to say, implicit diff isnt really 2u'ish at all...

I was referring to the point gradient formula.