quadratic intersecting circle: circle w/ center C(3,0) passes thru A(0,6); parabola passes thru 2 pts B, C, with...

N

nanase

Junior Member
Joined
Aug 8, 2019
Messages
243
Hello, I have been trying to solve this question, I need some help in finding the coordinates.
prelimA15 Q.jpeg
From the question, I realised that if AB and AC are perpendicular, their gradients are reciprocal of one another.
AB and AC are also radius of the circle. I also attempted to find the distance of AB and AC and equate them as they are equal.
Please correct and guide my steps, thank you.
prelimA15 working.jpeg
I ended up with fractional x and I'm not confident anymore with my steps, is this the right going?
 
I do realise I can probably set up a circle equation in the form (x-3)^2 + y^2 = 45.
But not sure if it will even more complicate the steps or not.
 
nanaseailie said:
Hello, I have been trying to solve this question, I need some help in finding the coordinates.
View attachment 37212
From the question, I realised that if AB and AC are perpendicular, their gradients are reciprocal of one another.
AB and AC are also radius of the circle. I also attempted to find the distance of AB and AC and equate them as they are equal.
Please correct and guide my steps, thank you.
View attachment 37214
I ended up with fractional x and I'm not confident anymore with my steps, is this the right going?
Click to expand...
Triangle ABC is a right isosceles triangle (45-45-90). You found the length of AC, what's the length of AB?

Next, set up a system of equations to find the coordinate of B.

Once you find the coordinate B, you can set up another system of equations to solve for b & c.

Lastly, knowing the equation of the parabola you can find its x-intercepts.

PS: Alternatively, you can use vector and dot products to find the coordinate of B if you don't want to work with a system of equations.
 
Last edited:
BigBeachBanana said:
Triangle ABC is a right isosceles triangle (45-45-90). You found the length of AC, what's the length of AB?

Next, set up a system of equations to find the coordinate of B.

Once you find the coordinate B, you can set up another system of equations to solve for b & c.

Lastly, knowing the equation of the parabola you can find its x-intercepts.

PS: Alternatively, you can use vector and dot products to find the coordinate of B if you don't want to work with a system of equations.
Click to expand...
Hi BBB,
It took me a while to try what you're suggesting and here we go.

As you suggested, from the top left I used the distance formula, and subs. the linear equation, I found x = 9. and y=3 as the coordinates of B.
Then I subs that into the quadratic equation.
Now I have two equations that are 9b+c=43.5 and 3b+c=4.5
Solving them simultaneously, I got b = 6.5 and c = -15. Thus I got my quadratic equation.
I then find the x-intercept and got two x values, that are x=15 and x=-2
Judging from the diagram, I opted for x=15, thus the x-intercept is (15,0)
.
.
But the answer key said it should be (10,0).
would you mind telling me what went wrong in my process? Thank you
 

Attachments

  • prelimA15 working2.jpeg
    prelimA15 working2.jpeg
    84.8 KB · Views: 3
Last edited:
nanaseailie said:
Hi BBB,
It took me a while to try what you're suggesting and here we go.
View attachment 37223
As you suggested, from the top left I used the distance formula, and subs. the linear equation, I found x = 9. and y=3 as the coordinates of B.
Then I subs that into the quadratic equation.
Now I have two equations that are 9b+c=43.5 and 3b+c=4.5
Solving them simultaneously, I got b = 6.5 and c = -15. Thus I got my quadratic equation.
I then find the x-intercept and got two x values, that are x=15 and x=-2
Judging from the diagram, I opted for (15,0)
.
.
But the answer key said it should be (10,0).
would you mind telling me what went wrong in my process? Thank you
Click to expand...
Solve your quadratic again. You already know one of the roots is (3,0) so your answers cannot be right.
 
Last edited:
BigBeachBanana said:
S

Solve your quadratic again. You already know one of the roots is (3,0) so your answers cannot be right.
Click to expand...
which part of the quadratic?
from x^2 - 6x - 27 = 0, I got x = 9 and x = -3
I have rechecked this part, and I think all my algebraic working is correct. Is there a wrong substitution?
 
OH MY GOOOOODDDDDDD
sorry and thank you!!!!
(x-3)(x-10)
thank you so much BBB, I love you to the moon and back
 
You must log in or register to reply here.
Top