Quadratics Help

**supersaiyan** · July 16th 2007, 05:58 PM

k i am a grade 11 student and need help with some questions

These question are from Grade 11 Nelson Book (Canadian) From a quadratic section

1a) The line y = x - 1 intersects the circle X^2 + y^2 = 25 at two points. This line is called a secant. Find the coordinates of the two points of intersection.

1b) For what value(s) of k will the line y = x + k be a tangent to the circle x^2 + y^2 = 25? A tangent is a line that touches a circle exactly one point.

2) vitaly and jen have 24 m of fencing to enclose a vegetable garden at teh back of their house. What are the dimensions of the largest rectangular garden they could enclose with the length of fencing?
Pic

**Jonboy** · July 16th 2007, 06:48 PM

I'll help with the first. So we have the equations:

(I).. $\;y\,=\,x\,-\,1$

(II).. $\;x^2\,+\,y^2\,=\,25$

To find the point of intersection with algebra, we just solve the system for x and y.

So let's substitute the value of y in (I) into (II):

$x^2\,+\,(x\,-\,1)^2\,=\,25$

Expand $(x\,-\,1)^2$ :.. $x^2\,+\,x^2\,-\,2x\,+\,1\,=\,25$

Simplify:.. $2x^2\,-\,2x\,-\,12=\,0\;\Rightarrow\;x^2\,-\,x\,-\,12\,=\,0$

Factor:.. $(x\,-\,4)(x\,+\,3)\,=\,0\;\Rightarrow\;x\,=\,4\,,-\,3$

Now substitute our value of x into (I):

For the x value of 4:.. $y\,=\,(4)\,-\,1\,=\,3$

So we have the point:.. $(4\,,\,3)$

For the x value of -3, substituting it into (I), we get -4.

So we also have the other point $(-\,3\,,\,-\,4)$

The graph agrees with my answer:

**supersaiyan** · July 16th 2007, 06:57 PM

Wow Thanks alot for your help
That was really fast

Thanks again

**red_dog** · July 16th 2007, 08:57 PM

1b)
The system $\displaystyle \left\{\begin{array}{ll}x^2+y^2=25\\y=x+k\end{arra y}\right.$ must have a single solution.
Replacing $y$ in the first equation we obtain $2x^2+2kx+k^2-25=0$ .
This equation must have equal roots, so $\triangle =b^2-4ac=0\Leftrightarrow k^2=50\Rightarrow k=\pm 5\sqrt{2}$

1c)
Let $x,y$ be the length sides of the rectangle.
So we have $2x+2y=24\Rightarrow x+y=12$ . (1)
The area of the rectangle is $A=xy$ .
Replacing $y$ from (1) in the area we obtain $A=x(12-x)$ .
We can consider the area as a function of variable $x$ and we must find the maximum of this function.
The maximum of a quadratic function $f(x)=ax^2+bx+c,a<0$ is $\displaystyle -\frac{\triangle}{4a}$ , where $\triangle =b^2-4ac$ .
In this case $f(x)=-x^2+12x$ and the maximum is 36.
This maximum is obtained when $x=y=6$ , so the rectangle is a square.

**supersaiyan** · July 17th 2007, 04:27 AM

Thanks alot for the sloution

I love this forum

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