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Math Help - Quadratics Help

  1. #1
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    Quadratics Help

    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
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  2. #2
    Member Jonboy's Avatar
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    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:

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  3. #3
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    Wow Thanks alot for your help
    That was really fast

    Thanks again
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  4. #4
    MHF Contributor red_dog's Avatar
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    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.
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  5. #5
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    Thanks alot for the sloution

    I love this forum
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