Title: Conics
Course: Demonstrating STACK (v4.4.7)

Find the centre and radius of the circle defined by \( [value]=0\).

Centre is \( ( \)  \( , \)  \( ) \)

Radius is
Feedback: The equation of a circle with centre \((a,b)\) and radius \(r\) is \[ (x-a)^2+(y-b)^2=r^2.\] To rewrite the equation \([value]\) in standard form we need to complete the square for each variable. Note that \[ [value] = [value]\] and \[ [value] = [value].\] Combinging these results we have the standard form as \[ [value]=0 \] You can check this by expanding out. From this we can read the coordinates of the centre as \( ([value], [value])\) and the radius is \( [value] \).

Find the equation of the circle with centre \( ([value],[value])\) and which passes through \(([value],[value])\).
Feedback: The equation of a circle with centre \((a,b)\) and radius \(r\) is \[ (x-a)^2+(y-b)^2=r^2.\] The centre is \( ([value],[value])\) the circle passes through \(([value],[value])\) so therefore the radius is \[ \sqrt{ ([value]-[value])^2 + ([value]-[value])^2} = [value].\] Therefore the equation of the circle is \[ [value]. \]

Find the equation of the circle whose centre is \( ([value],[value]) \) and radius is \([value]\).
Feedback: The equation of a circle with centre \((a,b)\) and radius \(r\) is \[ (x-a)^2+(y-b)^2=r^2.\] In our case this is \[ [value].\]

Find the equations of the tangent and normal to the circle \[ [value]\] at the point \(([value],[value])\).

Tangent:

Normal:
Feedback: Tangents: Circles are a special case of the ellipse centred at \((x_c,y_c)\) \[ \frac{(x-x_c)^2}{a^2}+\frac{(y-y_c)^2}{b^2}=1.\] The tangent to this ellipse, through the point \((x',y')\), is given by \[ \frac{(x-x_c)(x'-x_c)}{a^2}+\frac{(y-y_c)(y'-y_c)}{b^2}=1.\] In our case we have \(a^2=b^2=[value]\) and \(x'=[value]\) and \(y'=[value]\). Using these values we have the tangent line as \[ [value]\] Expanding and rearranging give \[ [value]\] Normals: A line through \((x',y')\) with gradient \(m\) has equation \[ y=m(x-x')+y'.\] If a line has gradient \(m\) then the perpendicular line has gradient \(-1/m\). Our tangent line has gradient [value], so the normal line through \(([value],[value])\) has equation \[ y= [value] = [value].\]

Find the equations of the tangent and normal to the circle \[ [value]\] at the point \(([value],[value])\).

Tangent:

Normal:
Feedback: Tangents: Circles are a special case of the ellipse centred at \((x_c,y_c)\) \[ \frac{(x-x_c)^2}{a^2}+\frac{(y-y_c)^2}{b^2}=1.\] The tangent to this ellipse, through the point \((x',y')\), is given by \[ \frac{(x-x_c)(x'-x_c)}{a^2}+\frac{(y-y_c)(y'-y_c)}{b^2}=1.\] In our case we have \(a^2=b^2=[value]\) and the equation of our circle is \[ [value]\] Using \(x'=[value]\) and \(y'=[value]\) we have the tangent line as \[ [value]\] Expanding and rearranging give \[ [value]\] Normals: A line through \((x',y')\) with gradient \(m\) has equation \[ y=m(x-x')+y'.\] If a line has gradient \(m\) then the perpendicular line has gradient \(-1/m\). Our tangent line has gradient [value], so the normal line through \(([value],[value])\) has equation \[ y= [value] = [value].\]

The standard equation of an ellipse (after a shift) is of the form:

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 .\]

Find the equation in  standard (shifted) form of an ellipse with foci at  \(([value],[value])\) and \(([value],[value])\), and a semi-major axis \([value]\).

Give your answer as a list containing (in this order) the coordinates of the centre, \(x_c\) and \(y_c\), the value of \(a\) and the value of \(b\).

\([ x_c, y_c, a, b] \) =
Feedback: The centre is the midpoint of the foci \(([value],[value])\) and \(([value],[value])\), hence is given by the coordinates \(([value],[value])\). If the foci and centre are on a vertical line, the semi-major axis is \(b\). The distance from the foci to the centre is then \(f = e\cdot b\), which gives us the eccentricity \(e\). The semi-minor axis \(a\) can then be found from \(a^2 = b^2(1-e^2)\). If the foci and centre are on an horizontal line, the semi-major axis is \(a\). The distance from the foci to the centre is then \(f = e\cdot a\), which gives us the eccentricity \(e\). The semi-minor axis \(b\) can then be found from \(b^2 = a^2(1 -e^2)\). In this case, the centre is given by \(([value],[value])\), \(a = [value]\) and \(b = [value]\). The equation of this conic is given by: \[ \frac{([value])^2}{[value]^2} + \frac{([value])^2}{[value]^2} = 1.\]

The standard equation of a hyperbola (after a shift) is of one of these two formats:

 \[ (1) \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, \quad   (2) \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 .\]

Find the equation in  standard (shifted) form of a hyperbola with centre \(([value],[value])\), a focus at \(([value],[value])\) and a directrix at \([value] = \frac[value][value].\)

Give your answer as a list containing (in this order) the format, the value of \(a\) and the value of \(b\).

\( [ format, a, b] \) =
Feedback: If the directrix is a vertical line (\(x=\alpha\)) then the hyperbola with centre at \((0,0)\) will be in form (1), with the foci at \((a\cdot e,0)\) and \((-a\cdot e,0)\) and one directrix given by \(x = \frac{a}{e}\), and the other given by \(x=-\frac{a}{e}\). Hence \(a\) and \(e\) can be calculated and \(b\) is found using \(b^2 = a^2(e^2-1)\). If the directrix is an horizontal line (\(y=\alpha\)) then the hyperbola with centre at \((0,0)\) will be in form (2), with the foci at \((0,b\cdot e)\) and \((0,-b\cdot e)\) and one directrix given by \(y = \frac{b}{e}\), and the other given by \(y=-\frac{b}{e}\). Hence \(b\) and \(e\) can be calculated and \(b\) is found using \(a^2 = b^2(e^2-1)\). In this case, the appropriate form is (\([value]\)) and \(a=[value], b=[value]\). The equation of this conic is given by: \[ [value] \].

The following equation represents two straight lines. Determine the point of intersection of these two lines:
\[ [value] .\]
\([ x,\, y ]\) =
Feedback: The quadratic equation can be factorised as \[([value])([value]) = 0,\] so we need to find the solution of the system \[[value],\] \[[value].\] This gives \([x,y] = [value]\).

The equation given below represents a parabola: \[ [value]. \]
Find the coordinates of the vertex, \((x_c,y_c)\), the coordinates of the focus, \((x_f,y_f)\) and the equation of the directrix.

Give your answer as a list containing (in this order) the coordinates of the vertex, the coordinates of the focus and the equation of the directrix.
\( [ x_c, y_c, x_f, y_f, eq ] \) =
Feedback: If the equation can be written in the form \[ (y - y_c) = a\cdot(x-x_c)^2,\] then the vertex is given by \((x_c,y_c)\), the focus is given by \((x_c, y_c +\frac{1}{4a})\), and the directrix by \(y = y_c - \frac{1}{4a}\). Or, if it can be written in the form \[ (x - x_c) = a\cdot(y-y_c)^2,\] then the vertex is given by \((x_c,y_c)\), the focus is given by \((x_c + \frac{1}{4a}, y_c)\), and the directrix by \(x = x_c - \frac{1}{4a}\). The equation \([value]\) can be written as: \[ [value][value] = ([value]) . \] Giving, \(a = [value]\), the coordinates of the vertex are \(([value],[value])\), the coordinates of the focus are \(([value],[value])\) and the directrix is \([value]\).

The standard equation of a conic takes one of the following five formats:

 \[ (1) \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \mbox{(ellipse)},\]

 \[ (2) \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1,\quad (3) \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1, \mbox{(hyperbola)},\]

 \[ (4) y = a x^2,\quad (5) x = a y^2, \mbox{(parabola)}.\]

Reduce the following quadratic equation of a conic to one of these standard forms:

 \[ [value] . \]

Give your answer as a list containing (in this order) the format number, the value of \(a\) and the value of \(b\).  In the case of a parabola, enter 0 for \(b\).

\([format,a,b]\) =
Feedback: Completing the squares in \(x\) and \(y\), the equation \([value]\) can be written as \[ [value].\] Hence this is in format \(([value])\) with \(a = [value]\) and \(b = [value]\). The correct answer is therefore \([value]\).

The following equation represents two straight lines. Determine the equation of each of the two lines:
\[ [value] .\]
Enter your answer as a set: \(\{a x + b y + c = 0,\, d x + e y + f = 0 \}\) =
Feedback: The quadratic equation \([value]\) can be factorised as \[([value])([value]) = 0,\] so the solution is \([value]\).