Rectangular hyperbola 1999 CSSA (1 Viewer)

[asianese]

The hyperbola touches the circle at the point Q.
i) Show this information on a sketch.
ii) Show that if is a repeated root of the polynomial equation then is also a root of
iii) Deduce that the equation has a repeated root , and two non-real complex roots.
iv) Find the values and the remaining roots of the equation
v) Find the equation of the common tangent to the hyperbola and the circle at Q.

I'm having trouble with iii)...Do I sub in c^4? I get it equalling 0 which is not good. Any tips/hints would be very much appreciated.

Thankyou!

 

[funnytomato]

is it just me , or you can't really see the question clearly?

 

[4025808]

right click, open the image in a new tab. then you'll be able to see.

 

[4025808]

Did you do this:

xy = c^2
x^2*y^2 = c^4, by squaring both sides
y^2 = c^4 / x^2
and sub y^2 into the equation of the circle, giving you the result of part iii)

 

[funnytomato]

solve simlutaneouly , as suggested by 4025808
you'd obtain the quartic in the question

circle TOUCHES hyperbola implies that there is a real solution that's a DOUBLE root
EDIT: it's doesn't , see #11

then from your graph, you can see that point must the be ONLY place where they intersect
hence the remaining 2 roots to the quartic must be complex

 

[bleakarcher]

solve simlutaneouly , as suggested by 4025808
you'd obtain the quartic in the question

circle TOUCHES hyperbola implies that there is a real solution that's a DOUBLE root

then from your graph, you can see that point must the be ONLY place where they intersect
hence the remaining 2 roots to the quartic must be complex

Why is it that if two curves touch that the x-coordinate of their point of intersection is a double root of their simultaneous equation?

 

[asianese]

OH wow I just made it way more complicated it for myself when needed...

@Bleakarcher: if you have say a parabola y=x^2+1, and the line y-1=0 (y=1). Solving these two simultaneously gives x^2+1=1, x^2=0. The line and the parabola touches twice, so there is a double root there, namely x. It's part of the discriminant theory.

 

[funnytomato]

the black one is the actual graph

if it's a single root(red curve), then it must intersect y=0 again,
giving us another real solution, condradicting the fact that there's only 1 point where they touch

 

[funnytomato]

OH wow I just made it way more complicated it for myself when needed...

@Bleakarcher: if you have say a parabola y=x^2+1, and the line y-1=0 (y=1). Solving these two simultaneously gives x^2+1=1, x^2=0. The line and the parabola touches twice, so there is a double root there, namely x. It's part of the discriminant theory.

but I think you'll need to justify why it's only a double root
not triple, or quadruple
[find P'(x) , P''(x), P'''(x) ]

PS: it could only be either double or quadruple, since they're touching

 

[4025808]

okay, so for iii), what we could do is this:

differentiate the main equation from iii), and we should get stationary points of x and x = 3/2. We find out that x = 0 is a horizontal point of inflexion. But we need to make a repeated root for when x = 3/2. So this means x = 3/2 MUST be a minimum turning point which is also = 0.
we sub in those points to find the y points of the graph y = x^2 (x-1)^2 + c^4 -x^2
when we sub x = 0, we get y = c^4
when we sub x = 3/2, we get y = -27/16 + c^4

but as we said before, the hyperbola is touching the circle, meaning that it is a double root. Therefore we want -27/16 + c^4 = 0, meaning c^4 = 27/16


iv)

From the above, we know that the beta is the repeated root, so beta = 3/2. We also know from above, c^4 = 27/16, but c^2 is always greater than 0, hence c^2 = 3 sqrt 3 / 4.

We use long division from the equation in part iii) and we should get complex roots as conjugates. Note that we divide by (2x-3)^2.

v)
We should obtain the equation of the hyperbola as xy = 3sqrt3 / 4

Sub x = 3/2, get y = sqrt3 /2

we find the equation of the tangent by differentiating the hyperbola with respect to x, then sub in x = 3/2 to get the gradient. from there, find the equation of the tangent to the hyperbola at x = 3/2 (should be easy).

FML my posting thingo has no latex

 

[funnytomato]

Why is it that if two curves touch that the x-coordinate of their point of intersection is a double root of their simultaneous equation?

thanks for reminding me
it doesn't always have to be a double root, but the multiplicity must be even
for example, f(x)=x^4

but in this case, it just happens to be a double root (since quadruple root doesn't work, and it's a quartic)

 

[asianese]

I think I solved it - anyone want to confirm?

iii) Solving simultaneously - deduce that Beta is a double root as it touches twice on the graph and no where else, so the other two roots must be complex.

iv) Using (ii), form and equation such that and so . Clearly x=0 is not the double root as at x=0, xy=c^2 is undefined! so the double root must be . So

Now from , put x=3/2, so that So the point of intersection is . Also, I'm not sure how to find the other two complex roots...I wolframalphaed it but it didn't give me the steps.

v) Using xy=c^2, . Equation of common tangent cbf with the tidying up.

 

[asianese]

LOL AWKWARD. I spent ages typing it out and we'd done the same thing 4025808 hahah

And thanks for the complex roots tip

 

[4025808]

I think I solved it - anyone want to confirm?

iii) Solving simultaneously - deduce that Beta is a double root as it touches twice on the graph and no where else, so the other two roots must be complex.

iv) Using (ii), form and equation such that and so . Clearly x=0 is not the double root as at x=0, xy=c^2 is undefined! so the double root must be . So

Now from , put x=3/2, so that So the point of intersection is . Also, I'm not sure how to find the other two complex roots...I wolframalphaed it but it didn't give me the steps.

v) Using xy=c^2, . Equation of common tangent cbf with the tidying up.

Looks all correct to me, except for iv), you have to say that c^4 = 27/16, c^2 > 0, therefore c^2 = 3sqrt3 /4

 

[asianese]

Ok thanks a lot!!

 

[funnytomato]

I think I solved it - anyone want to confirm?

iii) Solving simultaneously - deduce that Beta is a double root as it touches twice on the graph and no where else, so the other two roots must be complex.

iv) Using (ii), form and equation such that and so . Clearly x=0 is not the double root as at x=0, xy=c^2 is undefined! so the double root must be . So

Now from , put x=3/2, so that So the point of intersection is . Also, I'm not sure how to find the other two complex roots...I wolframalphaed it but it didn't give me the steps.

v) Using xy=c^2, . Equation of common tangent cbf with the tidying up.

since x=3/2 is a double root, then P(3/2)= (3/2)^4-2(3/2)^3+c^4=0 ,we'd get c^4=27/16
[I expanded everything, P(x)=x^4-2x^3+c^4]

then P(x)=x^4-2x^3+27/16 has a double root x=3/2
let the other 2 roots be a and b

a+b+3/2+3/2=2 ==> a+b=-1
ab(3/2)*(3/2)=27/16 ==> ab=3/4

so a, b are the solutions of
x^2+x+3/4=0
and solve to obtain the other 2 roots, which are complex conjugates since coeff. are real


PS: probably need to explain a bit more in part (iii)
and I think the rest of it is correct

 

[4025808]

since x=3/2 is a double root, then P(3/2)= (3/2)^4-2(3/2)^3+c^4=0 [I expanded everything, P(x)=x^4-2x^3+c^4]
we'd get c^4=27/16

then P(x)=x^4-2x^3+27/16 has a double root x=3/2
let the other 2 roots be a and b

a+b+3/2+3/2=2 ==> a+b=-1
ab(3/2)*(3/2)=27/16 ==> ab=3/4

so a, b are the solutions of
x^2+x+3/4=0
and solve to obtain the other 2 roots, which are complex conjugates since coeff. are real

I'd personally use long division in this section, but good work (Y)