Find the vertex,focus,axis,directrix and latus-rectum of the following parabolas
\(\\(i)~y^2=8x\\(ii)~4x^2+y=0\\(iii)~y^2-4y-3x+1=0\\(iv)~y^2-4y+4x=0\\(v)~~y^2+4x+4y-3=0\\(vi)~y^2=8x+8y\\(vii)~4(y-1)^2=-7(x-3)\\(viii)~y^2=5x-4y-9\\(ix)~x^2+y=6x-14\)
Find the vertex,focus,axis,directrix and latus-rectum of the following parabolas
\(\\(i)~y^2=8x\\(ii)~4x^2+y=0\\(iii)~y^2-4y-3x+1=0\\(iv)~y^2-4y+4x=0\\(v)~~y^2+4x+4y-3=0\\(vi)~y^2=8x+8y\\(vii)~4(y-1)^2=-7(x-3)\\(viii)~y^2=5x-4y-9\\(ix)~x^2+y=6x-14\)
\(\\(i)~\text{The given parabola}~y^2=8x~\text{is of the form}~y^2=4ax~where~4a=8\\~~~~~~\Rightarrow ~a=\frac{8}{4}=2\\~~~~~~\text{Vertex: The coordinates of the vertex are (0,0)}\\~~~~~~~~\text{Focus: The coordinates of the focus are (2,0)}\\~~~~~~~~\text{Axes: The equation of the axis is y=0}\\~~~~~~~~\text{Directrix:The equation of the directrix is x=-2}\\~~~~~~~~\text{Latus-rectum:The length of the latus-rectum}\\~~~~~~~~=4a=4 \times 2 =8 ~units.\\(ii)~\text{In the given parabola,}~a=\frac{1}{16}\\~~~~~~~~\text{Focus}~\left ( 0,-\frac{1}{16} \right )\\~~~~~~~~\text{vertex(0,0)}\\~~~~~~~~Directrix,y=\frac{1}{16}\\~~~~~~~~axis,x=0\\~~~~~~~~LR=\frac{1}{4}~units.\\(iii)~\text{The given equation is}\\~~~~~~~y^2-4y-3x+1=0\\~~~~~~~\Rightarrow y^2-4y=3x-1\\~~~~~~~\Rightarrow y^2-4y+4=3x-1+4\\~~~~~~~\Rightarrow y^2-4y+(2)^2=3x+3\\~~~~~~~\Rightarrow(y-2)^2=3(x+1)~~~~~~~~...(i)\\~~~~~~~~~~~~~~\text{Shifting the origin to the point (-1,2) without rotating the axes and denoting the new coordinates with respect to these axes by X and Y,we have}\\~~~~~~~~~~~~~~x=X-1,y=Y+2~~~~~~~~...(ii)\\~~~~~~~~~~~~~~\text{Using these relations equation~(i),reduces to}\\~~~~~~~~~~~~~~y^2=3X~~~~~~~~...(iii)\\~~~~~~~~~~~~~~~\text{This is of the form}~Y^2=4aX\\~~~~~~~~~~~~~~\text{On~comparing we get,}\\~~~~~~~~~~~~~~4a=3 \\~~~~~~~~~~~~~~~~~\ \Rightarrow ~~~a=\frac{3}{4}\\~~~~~~~~~~~~~~~~~\text{Now,}\\~~~~~~~~~~~~~~~~~\text{Vertex:the coordinates of the vertex with}\\~~~~~~~~~~~~~~~~~\text{respect to new axes are}~(X=0,Y=0)\\~~~~~~~~~~~~~~~~~\text{So,coordinates of the vertex with respect to old axes are}~(-1,2)\\~~~~~~~~~~~~~~~~~\text{Focus:The coordinates of the focus with respect to new axes are }~\left ( X=\frac{3}{4},Y=0 \right )\\~~~~~~~~~~~~~~~~~\text{Putting}~X=\frac{3}{4}~and~Y=0~in~equation~(ii),we~get\\~~~~~~~~~~~~~~~~~x=\frac{3}{4}-1~and~y=0+2\\~~~~~~~~~~~~~~~~~\Rightarrow ~x=\frac{-1}{4}~and~y=2\\~~~~~~~~~~~~~~~~~\therefore~~~\text{Coordinates of the focus with respect to old axes are}~\left ( \frac{-1}{4},2 \right )\\~~~~~~~~~~~~~~~~~\text{Axes: Equation of the axis of the parabola w.r.t. new axes is Y=0}\\~~~~~~~~~~~~~~~~~\Rightarrow y=2\\~~~~~~~~~~~~~~~~~\therefore~~~\text{equation of axis w.r.t. old axes is y=2}\\~~~~~~~~~~~\text{Directrix: Equation of the directrix of the parabola w.r.t new axes is}~~X=\frac{-3}{4}\\~~~~~~~~~~~~~~~~~\therefore~x=\frac{-3}{4}-1\\~~~~~~~~~~~~~~~~~\Rightarrow x=\frac{-7}{4}\\~~~~~~~~~~~~~~~~~\therefore~\text{Equation of the directrix of the parabola w.r.t. old axes is }~x=\frac{-7}{4}\\~~~~~~~~~~~~\text{Latus-rectum:The length of the latus-rectum }=4a\\~~~~~~~~~~~~~~~~~=4 \times \frac{3}{4}\\~~~~~~~~~~~~~~~~~=3~units\\(iv)~\text{The given equation is } ~y^2-4y+4x=0\\~~~~~~~~~~~~~~~\Rightarrow y^2-4y=-4x\\~~~~~~~~~~~~~~~\Rightarrow y^2-2xy \times 2+(2)^2=-4x+(2)^2\\~~~~~~~~~~~~~~~\Rightarrow (y-2)^2=-4x+4\\~~~~~~~~~~~~~~~\Rightarrow (y-2)^2=-4(x-1)~~~~~~~~~~~~~~~...(i)\\~~~~~~~~~~~~~~~\text{Shifting the origin to the point (1,2) without rotating the axes and denoting the new coordinates with respect to these axes by X and Y,we have}\\~~~~~~~~~~~~~~~x=X+1,y=Y+2~~~~~~~~~~~~~~~...(ii)\\~~~~~~~~~~~~~~~\text{Using these relations equation (i),reduces to }\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~Y^2=-4X~~~~~~~~~~~~~~~...(iii)\\~~~~~~~~~~~~~~~\text{This is of the form }~Y^2=-4aX\\~~~~~~~~~~~~~~~\text{on comparing,we get,a=1}\\~~~~~~~~~~~~~~~Now,\\~~~~~~~~~~~~~~~\text{Vertex: The coordinates of the vertex w.r.t. to new axes are (X=0,Y=0)}\\~~~~~~~~~~~~~~~\therefore~x=0+1,y=0+2\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[Using ~equation~ (ii)\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~x=1,y=2\\~~~~~~~~~~~~~~~\therefore~\text{Coordinates of the vertex w.r.t. old axes are (1,2)}\\~~~~~~~~~~~~~~~\text{Focus:The coordinates of the focus with respect to new axes are (X=1,Y=0)}\\~~~~~~~~~~~~~~~\text{Putting X=-1 and Y=0 in equation (ii),we get}\\~~~~~~~~~~~~~~~\therefore~y=0+2~~~~[Using~equation~(ii)]\\\\~~~~~~~~~~~~~~~\Rightarrow~y=2\\~~~~~~~~~~~~~~~\therefore~\text{equation of axis w.r.t. old axes is y=2}\\~~~~~~~~~~~~~~~\text{Directrix:Equation of the directirx of the parabola w.r.t. new axes is X=1}\\~~~~~~~~~~~~~~~\therefore~x=1+1~~~~~~~~~~~~~~~~~[Using~equation~(ii)]\\~~~~~~~~~~~~~~~~~~\Rightarrow~x=2\\~~~~~~~~~~~~~~~~~~\therefore~\text{Equation of the directrix of the parabola w.r.t. old axes is x=2}\\~~~~~~~~~~~~~~~~~~\text{Latus-rectum:The length of the latus-rectum =4a}\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=4 \times 1\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=4\)
\((v)~\text{The given equation is}~~y^2+4x+4y-3=0\\~~~~~~~~~~~~~~~~~\Rightarrow y^2+4y=-4x+3\\~~~~~~~~~~~~~~~~~\Rightarrow~~y^2+2 \times y \times 2 +(2)^2=-4x+3+2^2\\~~~~~~~~~~~~~~~~~\Rightarrow~~(y+2)^2=4x+3+4\\~~~~~~~~~~~~~~~~~\Rightarrow~~(y+2)^2=-4x+7\\~~~~~~~~~~~~~~~~~\Rightarrow~~(y+2)^2=-4\left ( x-\frac{7}{4} \right )~~~~~~~~~...(i)\\~~~~~~~~~~~~~~~~~\text{Shifting the origin to the point}~\left ( \frac{7}{4},-2 \right )~\text{without rotating the axes and denoting the new coordinates with respect to these axes by X and Y,we have}\\~~~~~~~~~~~~~~~~~x=X+\frac{7}{4},y=Y-2~~~~~~~~~...(ii)\\~~~~~~~~~~~~~~~~~\text{Using these relations equation (i),reduces to }~y^2=-4x~~~~~~~~~~~~~~~~~~~~~...(iii)\\~~~~~~~~~~~~~~~~~\text{This is of the form }~Y^2=-4aX~\text{on comparing,we get a=1}\\~~~~~~~~~~~~~~~~~~~~Now,\\~~~~~~~~~~~~~~~~~\text{Vertex:The coordinates of the vertex w.r.t. new axes are (X=0,Y=0)}\\~~~~~~~~~~~~~~~~~\therefore~x=0+\frac{7}{4},y=0-2~~~~~~~\text{[Using (ii)]}\\~~~~~~~~~~~~~~~~~\Rightarrow~~~~~~x=\frac{7}{4},y=-2\\~~~~~~~~~~~~~~~~~\therefore~~~~~~\text{Coordinates of the vertex w.r.t. old axes are }~~\left ( \frac{7}{4},-2 \right )\\~~~~~~~~~~~~~~~~~\text{Axis:Equation of the axis of the parabola w.r.t new axes is}\\~~~~~~~~~~~~~~~~~~~~~~~~Y=0\\~~~~~~~~~~~~~~~~~~~~~\therefore~~~~~y=0-2~~~~~~~~~\text{[Using equation (ii)]}\\~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow ~~~y=-2\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~~~\text{equation of the w.r.t old axes is y+2=0}\\~~~~~~~~~~~~~~~~~~~~~~~~\text{Directrix:Equation of the directrix of the parabola w.r.t new axes is}~~~~X=1\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~x=1+\frac{7}{4}~~~~~~~~~~~\text{[Using equation (ii)]}\\~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow ~~~~~~x=\frac{11}{4}\\~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow ~~~4x=11.\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~~~\text{Equation of the directrix of the parabola w.r.t. old axes is 4x=11}\\~~~~~~~~~~~~~~~~~~~~~~~~\text{Latus-rectum:The length of the latus-rectum =4a}\\~~~~~~~~~~~~~~~~~~~~~~~~=4 \times1\\~~~~~~~~~~~~~~~~~~~~~~~~=4 ~units\\~~~~~~~~~~~~~~~~~\text{Axis:Equation of the axis of the parabola w.r.t new axes is}\\~~~~~~~~~~~~~~~~~~~~~~~~Y=0\\~~~~~~~~~~~~~~~~~~~~~\therefore~~~~~y=0-2~~~~~~~~~\text{[Using equation (ii)]}\\~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow ~~~y=-2\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~~~\text{equation of the w.r.t old axes is y+2=0}\\~~~~~~~~~~~~~~~~~~~~~~~~\text{Directrix:Equation of the directrix of the parabola w.r.t new axes is}~~~~X=1\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~x=1+\frac{7}{4}~~~~~~~~~~~\text{[Using equation (ii)]}\\~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow ~~~~~~x=\frac{11}{4}\\~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow ~~~4x=11.\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~~~\text{Equation of the directrix of the parabola w.r.t. old axes is 4x=11}\\~~~~~~~~~~~~~~~~~~~~~~~~\text{Latus-rectum:The length of the latus-rectum =4a}\\~~~~~~~~~~~~~~~~~~~~~~~~=4 \times1\\~~~~~~~~~~~~~~~~~~~~~~~~=4 ~units\)
\((vi)~\text{The given equation is}~~y^2=8x+8y\\~~~~~~~~~~~~~~~~~\Rightarrow y^2-8y=8x\\~~~~~~~~~~~~~~~~~\Rightarrow~~y^2-2 \times 4 \times y +16=8x+16\\~~~~~~~~~~~~~~~~~\Rightarrow~~(y-4)^2=8(x+2)~~~~~~~~~~~~~~~~~...(i)\\~~~~~~~~~~~~~~~~~\text{Shifting the origin to the point}~(2,-4)~\text{without rotating the axes and denoting the new coordinates with respect to these axes by X and Y,we have}\\~~~~~~~~~~~~~~~~~x=X-2,y=Y+4~~~~~~~~...(ii)\\~~~~~~~~~~~~~~~~~\text{Using these relations equation (i),reduces to }~Y^2=8x\\~~~~~~~~~~~~~~~~~\text{This is of the form }~Y^2=4aX~\text{on comparing,we get}\\~~~~~~~~~~~~~~~~~~~~~~~~~4a=8\\~~~~~~~~~~~~~~~~~~\Rightarrow ~~~~~~~a=2\\~~~~~~~~~~~~~~~~~~~Now,\\~~~~~~~~~~~~~~~~~\text{Vertex:The coordinates of the vertex w.r.t. new axes are (X=0,Y=0)}\\~~~~~~~~~~~~~~~~~\therefore~x=0-2,y=0+4~\\~~~~~~~~~~~~~~\therefore~~\text{Coordinates of the vertex w.r.t. old axes are (-2,4)}\\~~~~~~~~~~~~~~\text{Focus:The coordinates of the focus w.r.t. new axes are (x=2,Y=0)}\\~~~~~~~~~~~~~~\therefore~~x=2-2~and~y=0+4\\~~~~~~~~~~~~~~~~~~~~~~~\text{[Using equation (ii)]}\\\\~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow ~~~~~~x=0~and~y=4\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~~~\text{cooridnates of the focus w.r.t. old axes are (0,4) }\\\text{Axis:Equation of the axis of the parabola w.r.t new axes is}\\~~~~~~~~~~~~~~~~~~~~~~~~Y=0\\~~~~~~~~~~~~~~~~~~~~~\therefore~~~~~y=0+4~~~~~~~~~\text{[Using equation (ii)]}\\~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow ~~~y=4\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~~~\text{equation of the w.r.t old axes is y=4}\\~~~~~~~~~~~~~~~~~~~~~~~~\text{Directrix:Equation of the directrix of the parabola w.r.t new axes is}~~~~x=-2\\\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~x=-2-2~~~~~~~~~~~\text{[Using equation (ii)]}\\~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow ~~~~~~x=-4\\~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow ~~~x+4=0.\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~~~\text{Equation of the directrix of the parabola w.r.t. old axes is x+4=0}\\~~~~~~~~~~~~~~~~~~~~~~~~\text{Latus-rectum:The length of the latus-rectum =4a}\\~~~~~~~~~~~~~~~~~~~~~~~~=4 \times2\\~~~~~~~~~~~~~~~~~~~~~~~~=8~units\)
\((vii)~\text{The given equation is}~~4(y-1)^2=-7(x-3)\\~~~~~~~~~~~~~~~~~\Rightarrow (y-1)^2=\frac{-7}{4}(x-3)\\~~~~~~~~~~~~~~~~~\text{Shifting the origin to the point}~(3,1)~\text{without rotating the axes and denoting the new coordinates with respect to these axes by X and Y,we have}\\~~~~~~~~~~~~~~~~~x=X+3,y=Y+1~~~~~~~~...(ii)\\~~~~~~~~~~~~~~~~~\text{Using these relations equation (i),reduces to }~Y^2=\frac{-7}{4}X~~~~~~~~~~~~~~~~~~...(iii)\\~~~~~~~~~~~~~~~~~\text{This is of the form }~Y^2=-4aX~\text{on comparing,we get}\\~~~~~~~~~~~~~~~~~~~~~~~~~4a=\frac{7}{4}\\~~~~~~~~~~~~~~~~~~\Rightarrow ~~~~~~~a=\frac{7}{16}\\~~~~~~~~~~~~~~~~~~~Now,\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~\text{Vertex:The coordinates of the vertex w.r.t. new axes are (X=0,Y=0)}\\~~~~~~~~~~~~~~~~~\therefore~x=0+3,y=0+1~\\~~~~~~~~~~~~~~~~~~~~~~~\text{[Using equation (iii)]}\\~~~~~~~~~~~~~~\therefore~~\text{Coordinates of the vertex w.r.t. old axes are (3,1)}\\~~~~~~~~~~~~~~~~~~~~\text{Focus:The coordinates of the focus w.r.t. new axes are }~~~~~~\left ( x=-\frac{7}{16},y=0 \right )\\~~~~~~~~~~~~~~\therefore~~x=x=\frac{-7}{16}+3,y=0+1\\~~~~~~~~~~~~~~~~~~\Rightarrow ~~~x=\frac{41}{16},y=1\\\\~~~~~~~~~~~~~~~~~\therefore~~~~~~\text{Coordinates of the vertex w.r.t. old axes are }~~\left ( \frac{41}{16},1\right )\\~~~~~~~~~~~~~~~~~\text{Axis:Equation of the axis of the parabola w.r.t new axes is}\\~~~~~~~~~~~~~~~~~~~~~~~~Y=0\\~~~~~~~~~~~~~~~~~~~~~\Rightarrow ~~~~~y=0+1~~~~~~~~~\\~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow ~~~y=1\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~~~\text{equation of the w.r.t old axes is y=1}\\~~~~~~~~~~~~~~~~~~~~~~~~\text{Directrix:Equation of the directrix of the parabola w.r.t new axes is}~~~~Y=\frac{7}{16}\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~x=\frac{7}{16}+3\\~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow x=\frac{55}{16}\\~~~~~~~~~~~\therefore~~~~~\text{Equation of the directrix of the parabola w.r.t old axes is}~~~x=\frac{55}{16}\\~~~~~~~~~~~~~~~~~~~~~~~~\text{Latus-rectum:The length of the latus-rectum =4a}\\~~~~~~~~~~~~~~~~~~~~~~~~=4 \times \frac{7}{16}\\~~~~~~~~~~~~~~~~~~~~~~~~=\frac{7}{4}~units\)
\((viii)~\text{The given equation is}~~4(y-1)^2=-7(x-3)\\~~~~~~~~~~~~~~~~~\Rightarrow y^2+4y=5x-9\\~~~~~~~~~~~~~~~~~\Rightarrow~~y^2+4y+4=5x-9+4\\~~~~~~~~~~~~~~~~~\Rightarrow~~(y+2)^2=5x-5\\~~~~~~~~~~~~~~~~~\Rightarrow~~(y+2)^2=5(x-1)~~~~~~~~~~~~~~~~~...(i)\\~~~~~~~~~~~~~~~~~\text{Shifting the origin to the point}~(1,-2)~\text{without rotating the axes and denoting the new coordinates with respect to these axes by X and Y,we have}\\~~~~~~~~~~~~~~~~~x=X+1,y=Y-2~~~~~~~~...(ii)\\~~~~~~~~~~~~~~~~~\text{Using these relations equation (i),reduces to }~Y^2=5x\\~~~~~~~~~~~~~~~~~\text{This is of the form }~Y^2=4aX~\text{on comparing,we get}\\~~~~~~~~~~~~~~~~~~~~~~~~~4a=5\\~~~~~~~~~~~~~~~~~~\Rightarrow ~~~~~~~a=\frac{5}{4}\\~~~~~~~~~~~~~~~~~~~Now,\\~~~~~~~~~~~~~~~~~\text{Vertex:The coordinates of the vertex w.r.t. new axes are (X=0,Y=0)}\\~~~~~~~~~~~~~~~~~\therefore~x=0+1~,y=0-2~\\~~~~~~~~~~~~~~~~~~~~~~~\text{[Using equation (ii)]}\\~~~~~~~~~~~~~~\text{Focus:The coordinates of the focus w.r.t. new axes are }~~~\left ( X=\frac{5}{4},Y=0 \right )\\~~~~~~~~~~~~~~\therefore~~x=\frac{5}{4}+1~and~y=0-2\\~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow ~~x=\frac{9}{4},y=-2\\~~~~~~~~~~~~~~~~~\therefore~~~~~~\text{Coordinates of the vertex w.r.t. old axes are }~~\left ( \frac{9}{4},-2\right )\\~~~~~~~~~~~~~~~~~~~~~~~~\text{Axis:Equation of the axis of the parabola w.r.t new axes is}\\~~~~~~~~~~~~~~~~~~~~~~~~Y=0\\~~~~~~~~~~~~~~~~~~~~~\Rightarrow ~~~~~y=0-2~~~~~~~~~\\~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow ~~~y=-2\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~~~\text{equation of the w.r.t old axes is y=-2}\\~~~~~~~~~~~~~~~~~~~~~~~~\text{Directrix:Equation of the directrix of the parabola w.r.t new axes is}~~~~X=\frac{-5}{4}\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~x=\frac{-5}{4}+1\\~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow x=\frac{-1}{4}\\~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~~4x+1=0\\~~~~~~~~~~~\therefore~~~~~\text{Equation of the directrix of the parabola w.r.t old axes is}~~~4x+1=0\\~~~~~~~~~~~~~~~~~~~~~~~~\text{Latus-rectum:The length of the latus-rectum =4a}\\~~~~~~~~~~~~~~~~~~~~~~~~=4 \times \frac{5}{4}\\~~~~~~~~~~~~~~~~~~~~~~~~=5~units\)
\((ix)~\text{The given equation is}~~x^2+y=6x-14\\~~~~~~~~~~~~~~~~~\Rightarrow x^2-6x=y-14\\~~~~~~~~~~~~~~~~~\Rightarrow~~x^2-2 \times x \times 3 +9=-y-14+9\\~~~~~~~~~~~~~~~~~\Rightarrow~~(x-3)^2=-y-5\\~~~~~~~~~~~~~~~~~\Rightarrow~~(x-3)^2=-1(y+5)~~~~~~~~~~~~~~~~~...(i)\\~~~~~~~~~~~~~~~~~\text{Shifting the origin to the point}~(3,-5)~\text{without rotating the axes and denoting the new coordinates with respect to these axes by X and Y,we have}\\~~~~~~~~~~~~~~~~~x=X+3,y=Y-5~~~~~~~~...(ii)\\~~~~~~~~~~~~~~~~~\text{Using these relations equation (i),reduces to }~X^2=-y\\~~~~~~~~~~~~~~~~~\text{This is of the form }~X^2=-4aY~\text{on comparing,we get}\\~~~~~~~~~~~~~~~~~~~~~~~~~4a=1\\~~~~~~~~~~~~~~~~~~\Rightarrow ~~~~~~~a=\frac{1}{4}\\~~~~~~~~~~~~~~~~~~~Now,\\~~~~~~~~~~~~~~~~~\text{Vertex:The coordinates of the vertex w.r.t. new axes are (X=0,Y=0)}\\~~~~~~~~~~~~~~~~~\therefore~x=0+3~,y=0-5~\\~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow x=3,y=-5\\\\~~~~~~~~~~~~~~~~~\therefore~~~~~~\text{Coordinates of the vertex w.r.t. old axes are }~~(3,-5)\\~~~~~~~~~~~~~~~~~~~~~~~~\text{Focus: The coordinates of the focus w.r.t new axes are}~~~\left ( X=0,Y=\frac{-1}{4} \right )\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~~x=0+3,y=\frac{-1}{4}-5\\~~~~~~~~~~~~~~~~~~~~~\Rightarrow ~~~~~x=3,y=\frac{-21}{4}~~~~~~~~~\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~~~\text{coordinates of the w.r.t old axes }~are~\left ( 3,\frac{-21}{4} \right )\\~~~~~~~~~~~~~~~~~~~~~~~~\text{Axis: Equation of the axis of the parabola w.r.t new axes is}~~~~X=0\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~x=0+3\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~x=3\\~~~~~~~~~~~\therefore~~~~~\text{Equation of the directrix of the parabola w.r.t old axes is}~~~x=3\\~~~~~~~~~~~~~~~~~~~~~~~~\text{Directrix:Equation of the directrix of the parabola w.r.t. new axes is}\\~~~~~~~~~~~~~~~~~~~~~~~~Y=\frac{1}{4}\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~y=\frac{1}{4}-5\\~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~y=\frac{-19}{4}\\~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~4y+19=0\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~\text{Equation of the directrix of the parabola w.r.t old axes is 4y+19=0}\\~~~~~~~~~~~~~~~~~~~~~~~~\text{Latus-rectum:The length of the latus-rectum =4a}\\~~~~~~~~~~~~~~~~~~~~~~~~=4 \times \frac{1}{4}\\~~~~~~~~~~~~~~~~~~~~~~~~=1~units \)
\(\\(i)~\text{The given parabola}~y^2=8x~\text{is of the form}~y^2=4ax~where~4a=8\\~~~~~~\Rightarrow ~a=\frac{8}{4}=2\\~~~~~~\text{Vertex: The coordinates of the vertex are (0,0)}\\~~~~~~~~\text{Focus: The coordinates of the focus are (2,0)}\\~~~~~~~~\text{Axes: The equation of the axis is y=0}\\~~~~~~~~\text{Directrix:The equation of the directrix is x=-2}\\~~~~~~~~\text{Latus-rectum:The length of the latus-rectum}\\~~~~~~~~=4a=4 \times 2 =8 ~units.\\(ii)~\text{In the given parabola,}~a=\frac{1}{16}\\~~~~~~~~\text{Focus}~\left ( 0,-\frac{1}{16} \right )\\~~~~~~~~\text{vertex(0,0)}\\~~~~~~~~Directrix,y=\frac{1}{16}\\~~~~~~~~axis,x=0\\~~~~~~~~LR=\frac{1}{4}~units.\\(iii)~\text{The given equation is}\\~~~~~~~y^2-4y-3x+1=0\\~~~~~~~\Rightarrow y^2-4y=3x-1\\~~~~~~~\Rightarrow y^2-4y+4=3x-1+4\\~~~~~~~\Rightarrow y^2-4y+(2)^2=3x+3\\~~~~~~~\Rightarrow(y-2)^2=3(x+1)~~~~~~~~...(i)\\~~~~~~~~~~~~~~\text{Shifting the origin to the point (-1,2) without rotating the axes and denoting the new coordinates with respect to these axes by X and Y,we have}\\~~~~~~~~~~~~~~x=X-1,y=Y+2~~~~~~~~...(ii)\\~~~~~~~~~~~~~~\text{Using these relations equation~(i),reduces to}\\~~~~~~~~~~~~~~y^2=3X~~~~~~~~...(iii)\\~~~~~~~~~~~~~~~\text{This is of the form}~Y^2=4aX\\~~~~~~~~~~~~~~\text{On~comparing we get,}\\~~~~~~~~~~~~~~4a=3 \\~~~~~~~~~~~~~~~~~\ \Rightarrow ~~~a=\frac{3}{4}\\~~~~~~~~~~~~~~~~~\text{Now,}\\~~~~~~~~~~~~~~~~~\text{Vertex:the coordinates of the vertex with}\\~~~~~~~~~~~~~~~~~\text{respect to new axes are}~(X=0,Y=0)\\~~~~~~~~~~~~~~~~~\text{So,coordinates of the vertex with respect to old axes are}~(-1,2)\\~~~~~~~~~~~~~~~~~\text{Focus:The coordinates of the focus with respect to new axes are }~\left ( X=\frac{3}{4},Y=0 \right )\\~~~~~~~~~~~~~~~~~\text{Putting}~X=\frac{3}{4}~and~Y=0~in~equation~(ii),we~get\\~~~~~~~~~~~~~~~~~x=\frac{3}{4}-1~and~y=0+2\\~~~~~~~~~~~~~~~~~\Rightarrow ~x=\frac{-1}{4}~and~y=2\\~~~~~~~~~~~~~~~~~\therefore~~~\text{Coordinates of the focus with respect to old axes are}~\left ( \frac{-1}{4},2 \right )\\~~~~~~~~~~~~~~~~~\text{Axes: Equation of the axis of the parabola w.r.t. new axes is Y=0}\\~~~~~~~~~~~~~~~~~\Rightarrow y=2\\~~~~~~~~~~~~~~~~~\therefore~~~\text{equation of axis w.r.t. old axes is y=2}\\~~~~~~~~~~~\text{Directrix: Equation of the directrix of the parabola w.r.t new axes is}~~X=\frac{-3}{4}\\~~~~~~~~~~~~~~~~~\therefore~x=\frac{-3}{4}-1\\~~~~~~~~~~~~~~~~~\Rightarrow x=\frac{-7}{4}\\~~~~~~~~~~~~~~~~~\therefore~\text{Equation of the directrix of the parabola w.r.t. old axes is }~x=\frac{-7}{4}\\~~~~~~~~~~~~\text{Latus-rectum:The length of the latus-rectum }=4a\\~~~~~~~~~~~~~~~~~=4 \times \frac{3}{4}\\~~~~~~~~~~~~~~~~~=3~units\\(iv)~\text{The given equation is } ~y^2-4y+4x=0\\~~~~~~~~~~~~~~~\Rightarrow y^2-4y=-4x\\~~~~~~~~~~~~~~~\Rightarrow y^2-2xy \times 2+(2)^2=-4x+(2)^2\\~~~~~~~~~~~~~~~\Rightarrow (y-2)^2=-4x+4\\~~~~~~~~~~~~~~~\Rightarrow (y-2)^2=-4(x-1)~~~~~~~~~~~~~~~...(i)\\~~~~~~~~~~~~~~~\text{Shifting the origin to the point (1,2) without rotating the axes and denoting the new coordinates with respect to these axes by X and Y,we have}\\~~~~~~~~~~~~~~~x=X+1,y=Y+2~~~~~~~~~~~~~~~...(ii)\\~~~~~~~~~~~~~~~\text{Using these relations equation (i),reduces to }\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~Y^2=-4X~~~~~~~~~~~~~~~...(iii)\\~~~~~~~~~~~~~~~\text{This is of the form }~Y^2=-4aX\\~~~~~~~~~~~~~~~\text{on comparing,we get,a=1}\\~~~~~~~~~~~~~~~Now,\\~~~~~~~~~~~~~~~\text{Vertex: The coordinates of the vertex w.r.t. to new axes are (X=0,Y=0)}\\~~~~~~~~~~~~~~~\therefore~x=0+1,y=0+2\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[Using ~equation~ (ii)\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~x=1,y=2\\~~~~~~~~~~~~~~~\therefore~\text{Coordinates of the vertex w.r.t. old axes are (1,2)}\\~~~~~~~~~~~~~~~\text{Focus:The coordinates of the focus with respect to new axes are (X=1,Y=0)}\\~~~~~~~~~~~~~~~\text{Putting X=-1 and Y=0 in equation (ii),we get}\\~~~~~~~~~~~~~~~\therefore~y=0+2~~~~[Using~equation~(ii)]\\\\~~~~~~~~~~~~~~~\Rightarrow~y=2\\~~~~~~~~~~~~~~~\therefore~\text{equation of axis w.r.t. old axes is y=2}\\~~~~~~~~~~~~~~~\text{Directrix:Equation of the directirx of the parabola w.r.t. new axes is X=1}\\~~~~~~~~~~~~~~~\therefore~x=1+1~~~~~~~~~~~~~~~~~[Using~equation~(ii)]\\~~~~~~~~~~~~~~~~~~\Rightarrow~x=2\\~~~~~~~~~~~~~~~~~~\therefore~\text{Equation of the directrix of the parabola w.r.t. old axes is x=2}\\~~~~~~~~~~~~~~~~~~\text{Latus-rectum:The length of the latus-rectum =4a}\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=4 \times 1\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=4\)
\((v)~\text{The given equation is}~~y^2+4x+4y-3=0\\~~~~~~~~~~~~~~~~~\Rightarrow y^2+4y=-4x+3\\~~~~~~~~~~~~~~~~~\Rightarrow~~y^2+2 \times y \times 2 +(2)^2=-4x+3+2^2\\~~~~~~~~~~~~~~~~~\Rightarrow~~(y+2)^2=4x+3+4\\~~~~~~~~~~~~~~~~~\Rightarrow~~(y+2)^2=-4x+7\\~~~~~~~~~~~~~~~~~\Rightarrow~~(y+2)^2=-4\left ( x-\frac{7}{4} \right )~~~~~~~~~...(i)\\~~~~~~~~~~~~~~~~~\text{Shifting the origin to the point}~\left ( \frac{7}{4},-2 \right )~\text{without rotating the axes and denoting the new coordinates with respect to these axes by X and Y,we have}\\~~~~~~~~~~~~~~~~~x=X+\frac{7}{4},y=Y-2~~~~~~~~~...(ii)\\~~~~~~~~~~~~~~~~~\text{Using these relations equation (i),reduces to }~y^2=-4x~~~~~~~~~~~~~~~~~~~~~...(iii)\\~~~~~~~~~~~~~~~~~\text{This is of the form }~Y^2=-4aX~\text{on comparing,we get a=1}\\~~~~~~~~~~~~~~~~~~~~Now,\\~~~~~~~~~~~~~~~~~\text{Vertex:The coordinates of the vertex w.r.t. new axes are (X=0,Y=0)}\\~~~~~~~~~~~~~~~~~\therefore~x=0+\frac{7}{4},y=0-2~~~~~~~\text{[Using (ii)]}\\~~~~~~~~~~~~~~~~~\Rightarrow~~~~~~x=\frac{7}{4},y=-2\\~~~~~~~~~~~~~~~~~\therefore~~~~~~\text{Coordinates of the vertex w.r.t. old axes are }~~\left ( \frac{7}{4},-2 \right )\\~~~~~~~~~~~~~~~~~\text{Axis:Equation of the axis of the parabola w.r.t new axes is}\\~~~~~~~~~~~~~~~~~~~~~~~~Y=0\\~~~~~~~~~~~~~~~~~~~~~\therefore~~~~~y=0-2~~~~~~~~~\text{[Using equation (ii)]}\\~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow ~~~y=-2\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~~~\text{equation of the w.r.t old axes is y+2=0}\\~~~~~~~~~~~~~~~~~~~~~~~~\text{Directrix:Equation of the directrix of the parabola w.r.t new axes is}~~~~X=1\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~x=1+\frac{7}{4}~~~~~~~~~~~\text{[Using equation (ii)]}\\~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow ~~~~~~x=\frac{11}{4}\\~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow ~~~4x=11.\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~~~\text{Equation of the directrix of the parabola w.r.t. old axes is 4x=11}\\~~~~~~~~~~~~~~~~~~~~~~~~\text{Latus-rectum:The length of the latus-rectum =4a}\\~~~~~~~~~~~~~~~~~~~~~~~~=4 \times1\\~~~~~~~~~~~~~~~~~~~~~~~~=4 ~units\\~~~~~~~~~~~~~~~~~\text{Axis:Equation of the axis of the parabola w.r.t new axes is}\\~~~~~~~~~~~~~~~~~~~~~~~~Y=0\\~~~~~~~~~~~~~~~~~~~~~\therefore~~~~~y=0-2~~~~~~~~~\text{[Using equation (ii)]}\\~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow ~~~y=-2\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~~~\text{equation of the w.r.t old axes is y+2=0}\\~~~~~~~~~~~~~~~~~~~~~~~~\text{Directrix:Equation of the directrix of the parabola w.r.t new axes is}~~~~X=1\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~x=1+\frac{7}{4}~~~~~~~~~~~\text{[Using equation (ii)]}\\~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow ~~~~~~x=\frac{11}{4}\\~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow ~~~4x=11.\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~~~\text{Equation of the directrix of the parabola w.r.t. old axes is 4x=11}\\~~~~~~~~~~~~~~~~~~~~~~~~\text{Latus-rectum:The length of the latus-rectum =4a}\\~~~~~~~~~~~~~~~~~~~~~~~~=4 \times1\\~~~~~~~~~~~~~~~~~~~~~~~~=4 ~units\)
\((vi)~\text{The given equation is}~~y^2=8x+8y\\~~~~~~~~~~~~~~~~~\Rightarrow y^2-8y=8x\\~~~~~~~~~~~~~~~~~\Rightarrow~~y^2-2 \times 4 \times y +16=8x+16\\~~~~~~~~~~~~~~~~~\Rightarrow~~(y-4)^2=8(x+2)~~~~~~~~~~~~~~~~~...(i)\\~~~~~~~~~~~~~~~~~\text{Shifting the origin to the point}~(2,-4)~\text{without rotating the axes and denoting the new coordinates with respect to these axes by X and Y,we have}\\~~~~~~~~~~~~~~~~~x=X-2,y=Y+4~~~~~~~~...(ii)\\~~~~~~~~~~~~~~~~~\text{Using these relations equation (i),reduces to }~Y^2=8x\\~~~~~~~~~~~~~~~~~\text{This is of the form }~Y^2=4aX~\text{on comparing,we get}\\~~~~~~~~~~~~~~~~~~~~~~~~~4a=8\\~~~~~~~~~~~~~~~~~~\Rightarrow ~~~~~~~a=2\\~~~~~~~~~~~~~~~~~~~Now,\\~~~~~~~~~~~~~~~~~\text{Vertex:The coordinates of the vertex w.r.t. new axes are (X=0,Y=0)}\\~~~~~~~~~~~~~~~~~\therefore~x=0-2,y=0+4~\\~~~~~~~~~~~~~~\therefore~~\text{Coordinates of the vertex w.r.t. old axes are (-2,4)}\\~~~~~~~~~~~~~~\text{Focus:The coordinates of the focus w.r.t. new axes are (x=2,Y=0)}\\~~~~~~~~~~~~~~\therefore~~x=2-2~and~y=0+4\\~~~~~~~~~~~~~~~~~~~~~~~\text{[Using equation (ii)]}\\\\~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow ~~~~~~x=0~and~y=4\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~~~\text{cooridnates of the focus w.r.t. old axes are (0,4) }\\\text{Axis:Equation of the axis of the parabola w.r.t new axes is}\\~~~~~~~~~~~~~~~~~~~~~~~~Y=0\\~~~~~~~~~~~~~~~~~~~~~\therefore~~~~~y=0+4~~~~~~~~~\text{[Using equation (ii)]}\\~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow ~~~y=4\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~~~\text{equation of the w.r.t old axes is y=4}\\~~~~~~~~~~~~~~~~~~~~~~~~\text{Directrix:Equation of the directrix of the parabola w.r.t new axes is}~~~~x=-2\\\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~x=-2-2~~~~~~~~~~~\text{[Using equation (ii)]}\\~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow ~~~~~~x=-4\\~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow ~~~x+4=0.\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~~~\text{Equation of the directrix of the parabola w.r.t. old axes is x+4=0}\\~~~~~~~~~~~~~~~~~~~~~~~~\text{Latus-rectum:The length of the latus-rectum =4a}\\~~~~~~~~~~~~~~~~~~~~~~~~=4 \times2\\~~~~~~~~~~~~~~~~~~~~~~~~=8~units\)
\((vii)~\text{The given equation is}~~4(y-1)^2=-7(x-3)\\~~~~~~~~~~~~~~~~~\Rightarrow (y-1)^2=\frac{-7}{4}(x-3)\\~~~~~~~~~~~~~~~~~\text{Shifting the origin to the point}~(3,1)~\text{without rotating the axes and denoting the new coordinates with respect to these axes by X and Y,we have}\\~~~~~~~~~~~~~~~~~x=X+3,y=Y+1~~~~~~~~...(ii)\\~~~~~~~~~~~~~~~~~\text{Using these relations equation (i),reduces to }~Y^2=\frac{-7}{4}X~~~~~~~~~~~~~~~~~~...(iii)\\~~~~~~~~~~~~~~~~~\text{This is of the form }~Y^2=-4aX~\text{on comparing,we get}\\~~~~~~~~~~~~~~~~~~~~~~~~~4a=\frac{7}{4}\\~~~~~~~~~~~~~~~~~~\Rightarrow ~~~~~~~a=\frac{7}{16}\\~~~~~~~~~~~~~~~~~~~Now,\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~\text{Vertex:The coordinates of the vertex w.r.t. new axes are (X=0,Y=0)}\\~~~~~~~~~~~~~~~~~\therefore~x=0+3,y=0+1~\\~~~~~~~~~~~~~~~~~~~~~~~\text{[Using equation (iii)]}\\~~~~~~~~~~~~~~\therefore~~\text{Coordinates of the vertex w.r.t. old axes are (3,1)}\\~~~~~~~~~~~~~~~~~~~~\text{Focus:The coordinates of the focus w.r.t. new axes are }~~~~~~\left ( x=-\frac{7}{16},y=0 \right )\\~~~~~~~~~~~~~~\therefore~~x=x=\frac{-7}{16}+3,y=0+1\\~~~~~~~~~~~~~~~~~~\Rightarrow ~~~x=\frac{41}{16},y=1\\\\~~~~~~~~~~~~~~~~~\therefore~~~~~~\text{Coordinates of the vertex w.r.t. old axes are }~~\left ( \frac{41}{16},1\right )\\~~~~~~~~~~~~~~~~~\text{Axis:Equation of the axis of the parabola w.r.t new axes is}\\~~~~~~~~~~~~~~~~~~~~~~~~Y=0\\~~~~~~~~~~~~~~~~~~~~~\Rightarrow ~~~~~y=0+1~~~~~~~~~\\~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow ~~~y=1\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~~~\text{equation of the w.r.t old axes is y=1}\\~~~~~~~~~~~~~~~~~~~~~~~~\text{Directrix:Equation of the directrix of the parabola w.r.t new axes is}~~~~Y=\frac{7}{16}\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~x=\frac{7}{16}+3\\~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow x=\frac{55}{16}\\~~~~~~~~~~~\therefore~~~~~\text{Equation of the directrix of the parabola w.r.t old axes is}~~~x=\frac{55}{16}\\~~~~~~~~~~~~~~~~~~~~~~~~\text{Latus-rectum:The length of the latus-rectum =4a}\\~~~~~~~~~~~~~~~~~~~~~~~~=4 \times \frac{7}{16}\\~~~~~~~~~~~~~~~~~~~~~~~~=\frac{7}{4}~units\)
\((viii)~\text{The given equation is}~~4(y-1)^2=-7(x-3)\\~~~~~~~~~~~~~~~~~\Rightarrow y^2+4y=5x-9\\~~~~~~~~~~~~~~~~~\Rightarrow~~y^2+4y+4=5x-9+4\\~~~~~~~~~~~~~~~~~\Rightarrow~~(y+2)^2=5x-5\\~~~~~~~~~~~~~~~~~\Rightarrow~~(y+2)^2=5(x-1)~~~~~~~~~~~~~~~~~...(i)\\~~~~~~~~~~~~~~~~~\text{Shifting the origin to the point}~(1,-2)~\text{without rotating the axes and denoting the new coordinates with respect to these axes by X and Y,we have}\\~~~~~~~~~~~~~~~~~x=X+1,y=Y-2~~~~~~~~...(ii)\\~~~~~~~~~~~~~~~~~\text{Using these relations equation (i),reduces to }~Y^2=5x\\~~~~~~~~~~~~~~~~~\text{This is of the form }~Y^2=4aX~\text{on comparing,we get}\\~~~~~~~~~~~~~~~~~~~~~~~~~4a=5\\~~~~~~~~~~~~~~~~~~\Rightarrow ~~~~~~~a=\frac{5}{4}\\~~~~~~~~~~~~~~~~~~~Now,\\~~~~~~~~~~~~~~~~~\text{Vertex:The coordinates of the vertex w.r.t. new axes are (X=0,Y=0)}\\~~~~~~~~~~~~~~~~~\therefore~x=0+1~,y=0-2~\\~~~~~~~~~~~~~~~~~~~~~~~\text{[Using equation (ii)]}\\~~~~~~~~~~~~~~\text{Focus:The coordinates of the focus w.r.t. new axes are }~~~\left ( X=\frac{5}{4},Y=0 \right )\\~~~~~~~~~~~~~~\therefore~~x=\frac{5}{4}+1~and~y=0-2\\~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow ~~x=\frac{9}{4},y=-2\\~~~~~~~~~~~~~~~~~\therefore~~~~~~\text{Coordinates of the vertex w.r.t. old axes are }~~\left ( \frac{9}{4},-2\right )\\~~~~~~~~~~~~~~~~~~~~~~~~\text{Axis:Equation of the axis of the parabola w.r.t new axes is}\\~~~~~~~~~~~~~~~~~~~~~~~~Y=0\\~~~~~~~~~~~~~~~~~~~~~\Rightarrow ~~~~~y=0-2~~~~~~~~~\\~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow ~~~y=-2\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~~~\text{equation of the w.r.t old axes is y=-2}\\~~~~~~~~~~~~~~~~~~~~~~~~\text{Directrix:Equation of the directrix of the parabola w.r.t new axes is}~~~~X=\frac{-5}{4}\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~x=\frac{-5}{4}+1\\~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow x=\frac{-1}{4}\\~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~~4x+1=0\\~~~~~~~~~~~\therefore~~~~~\text{Equation of the directrix of the parabola w.r.t old axes is}~~~4x+1=0\\~~~~~~~~~~~~~~~~~~~~~~~~\text{Latus-rectum:The length of the latus-rectum =4a}\\~~~~~~~~~~~~~~~~~~~~~~~~=4 \times \frac{5}{4}\\~~~~~~~~~~~~~~~~~~~~~~~~=5~units\)
\((ix)~\text{The given equation is}~~x^2+y=6x-14\\~~~~~~~~~~~~~~~~~\Rightarrow x^2-6x=y-14\\~~~~~~~~~~~~~~~~~\Rightarrow~~x^2-2 \times x \times 3 +9=-y-14+9\\~~~~~~~~~~~~~~~~~\Rightarrow~~(x-3)^2=-y-5\\~~~~~~~~~~~~~~~~~\Rightarrow~~(x-3)^2=-1(y+5)~~~~~~~~~~~~~~~~~...(i)\\~~~~~~~~~~~~~~~~~\text{Shifting the origin to the point}~(3,-5)~\text{without rotating the axes and denoting the new coordinates with respect to these axes by X and Y,we have}\\~~~~~~~~~~~~~~~~~x=X+3,y=Y-5~~~~~~~~...(ii)\\~~~~~~~~~~~~~~~~~\text{Using these relations equation (i),reduces to }~X^2=-y\\~~~~~~~~~~~~~~~~~\text{This is of the form }~X^2=-4aY~\text{on comparing,we get}\\~~~~~~~~~~~~~~~~~~~~~~~~~4a=1\\~~~~~~~~~~~~~~~~~~\Rightarrow ~~~~~~~a=\frac{1}{4}\\~~~~~~~~~~~~~~~~~~~Now,\\~~~~~~~~~~~~~~~~~\text{Vertex:The coordinates of the vertex w.r.t. new axes are (X=0,Y=0)}\\~~~~~~~~~~~~~~~~~\therefore~x=0+3~,y=0-5~\\~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow x=3,y=-5\\\\~~~~~~~~~~~~~~~~~\therefore~~~~~~\text{Coordinates of the vertex w.r.t. old axes are }~~(3,-5)\\~~~~~~~~~~~~~~~~~~~~~~~~\text{Focus: The coordinates of the focus w.r.t new axes are}~~~\left ( X=0,Y=\frac{-1}{4} \right )\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~~x=0+3,y=\frac{-1}{4}-5\\~~~~~~~~~~~~~~~~~~~~~\Rightarrow ~~~~~x=3,y=\frac{-21}{4}~~~~~~~~~\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~~~\text{coordinates of the w.r.t old axes }~are~\left ( 3,\frac{-21}{4} \right )\\~~~~~~~~~~~~~~~~~~~~~~~~\text{Axis: Equation of the axis of the parabola w.r.t new axes is}~~~~X=0\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~x=0+3\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~x=3\\~~~~~~~~~~~\therefore~~~~~\text{Equation of the directrix of the parabola w.r.t old axes is}~~~x=3\\~~~~~~~~~~~~~~~~~~~~~~~~\text{Directrix:Equation of the directrix of the parabola w.r.t. new axes is}\\~~~~~~~~~~~~~~~~~~~~~~~~Y=\frac{1}{4}\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~y=\frac{1}{4}-5\\~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~y=\frac{-19}{4}\\~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~4y+19=0\\~~~~~~~~~~~~~~~~~~~~~~~~\therefore~\text{Equation of the directrix of the parabola w.r.t old axes is 4y+19=0}\\~~~~~~~~~~~~~~~~~~~~~~~~\text{Latus-rectum:The length of the latus-rectum =4a}\\~~~~~~~~~~~~~~~~~~~~~~~~=4 \times \frac{1}{4}\\~~~~~~~~~~~~~~~~~~~~~~~~=1~units \)
