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Thread: Question based on conics

  1. #1
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    Question based on conics

    The following is a 'multiple-option correct' question..

    If the straight line 3x+4y=24 intersects the axes at A and B and the straight line 4x+3y=24 at C and D, then the points A,B,C and D lie on:
    (a) a Circle
    (b) a Parabola
    (c) an Ellipse
    (d) a Hyperbola

    My attempt:

    I got the points A,B,C and D by solving the given equations.. which are (0,6) (0,8) (6,0) (8,0).
    I am definitely sure that the option (A) is right..
    Because i got a clear equation of a circle with centre at (7,7) and its radius being √50..


    But how do i proceed for the other conics.... Kindly help
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Perhaps you have covered the following result:

    The family of conics passing through $\displaystyle A,B,C,D$ is

    $\displaystyle \{\lambda \;\overline{AB}\cdot \overline{CD}+ \overline{AC} \cdot \overline{BD}=0\;:\lambda\in\mathbb{R}\}\cup\{\ove rline{AB}\cdot \overline{CD}=0\}$
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    I am sorry.. But, i didnt quite get the equation which you gave me.. :|
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    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by animesh271094 View Post
    I am sorry.. But, i didnt quite get the equation which you gave me.. :|

    In such a case:

    (a) You have proved that the four points lie on a circle.

    (b) The four points lie on $\displaystyle \overline{AC}\cdot \overline {BD}=0$ that is on a pair of parallel lines (degenerated parabola).

    (c) A circle is a particular case of an elipse.

    (d) The four points lie on $\displaystyle (3x+4y-24)(4x+3y-24)=0$ that is, on a pair of non parallel lines (degenerated hyperbola).
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  5. #5
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    Hello, animesh271094!

    $\displaystyle \text{If the straight line }3x+4y\,=\,24\text{ intersects the axes at }A\text{ and }B$
    $\displaystyle \text{and the straight line }4x+3y\,=\,24\text{ at }C\text{ and }D,$
    $\displaystyle \text{then the points }A,B,C,D\text{ lie on:}$

    . . $\displaystyle \text{(a) a Circle }\quad \text{(b) a Parabola }\quad \text{(c) an Ellipse } \quad \text{(d) a Hyperbola}$


    $\displaystyle \text{My attempt:}$

    $\displaystyle \text{I got the points }A,B,C,D\!:\;(0,6),\:(0,8),\:(6,0),\:(8,0).$
    $\displaystyle \text{I am d{e}finitely sure that the option (A) is right.}$
    $\displaystyle \text{Because I got an equation of a circle}$
    . . $\displaystyle \text{with centre at }(7,7)\text{ and radius }\sqrt{50}.$ . Right!

    $\displaystyle \text{But how do i proceed for the other conics?}$

    If we are allowed "tipped" conics, all four answers are correct.


    The general quadratic equation is:
    . . $\displaystyle \,Ax^2 + Bxy + Cy^2 + Dx + Ey + F \:=\:0$

    $\displaystyle \text{The discriminant is: }\,\Delta \,=\,B^2-4AC\quad\begin{Bmatrix}\Delta \:=\:0\!: & \text{parabola} \\ \Delta \:>\:0\!: & \text{hyperbola} \\ \Delta \:<\:0\!: & \text{ellipse} \end{Bmatrix}$

    Substitute the four points . . .


    $\displaystyle \begin{array}{cccccccc}(8,0)\!: & 64A + 8D + F &=& 0 & [1] \\
    (6,0)\!: & 36A + 6D + F &=& 0 & [2] \end{array}$

    Subtract: .$\displaystyle 28A + 2D \:=\:0 \quad\Rightarrow\quad D \:=\:\text{-}14A\;\;[3]$

    Substitute into [2]: .$\displaystyle 36A + 6(\text{-}14A) + F \:=\:0 \quad\Rightarrow\quad F \:=\:48A\;\;[4]$


    $\displaystyle \begin{array}{cccccccc}(0.8)\!: & 64C + 8E + F &=& 0 & [5] \\ (0,6)\!: & 36C + 6E + F &=& 0 & [6] \end{array}$

    Subtract: .$\displaystyle 28C + 2E \:=\:0 \quad\Rightarrow\quad E \:=\:\text{-}14C\;\;[7]$

    Substitute into [6]: .$\displaystyle 36C + 6(\text{-}14C) + F \:=\:0 \quad\Rightarrow\quad F \:=\:48C\;\;[8]$


    From [4] and [8]: .$\displaystyle A \:=\:C$
    . . and [7] becomes: .$\displaystyle E \:=\:\text{-}14A$


    The quadratic equation becomes:

    . . $\displaystyle Ax^2 + Bxy + Ay^2 - 14Ax - 14Ay + 48A \:=\:0$


    If $\displaystyle B = 0$, we have the circle that you found.

    Otherwise, the conics are all rotated $\displaystyle 45^o$ clockwise.


    The discriminant is: .$\displaystyle \Delta \:=\:B^2-4A^2$

    Select values for $\displaystyle \,A$ and $\displaystyle \,B$ and you can have any of the conic curves.

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