Explanation
The image of circle w.r.t same line means image of centre w.r.t that line without changing radius x2+y2−6x−4y+12=0 Centre =(3,2) Radius=1 Image of (3,2) w.r.t x+y−1=0 x−31=y−21=−2(3+2−1)11+12=−4 x=−1, y=−2 Then equation of image of circle is (x+1)2+(y+2)2=(1)2 ⇒x2+y2+2x+4y+4=0
Given equation is x^{2}+y^{2}=2ax
Centre =(a,0) and radius =a
Therefore, equation of line having slope = \dfrac {-1}{2} and passing through (4,0) y=\dfrac {-1}{2}(x-a) Line AB: 2y+x=a -(1) y=\dfrac {a+x}{2} d=\left | \dfrac {a}{\sqrt{5}} \right | So, area of \triangle AOB =\dfrac {1}{2}\times\ AB\ \times (perpendicular distance of Q from AB ) =\dfrac {1}{2}\times 2a\times \dfrac {a}{\sqrt{5}} =\dfrac {a^{2}}{\sqrt{5}}
d=\left | \dfrac{C_{1}-C_{2}}{\sqrt{a^{2}+b^{2}}} \right |where d is the distance between lines whose equations are ax+by+C_{1}=0 & ax+by+C_{2}=0d=\left | \dfrac{27-2}{\sqrt{4^{2}+3^2}} \right |=5d=5If the distance between vertex and latus rectum=distance of vertex from directrix=a
then d=2a=5\Rightarrow Length of latus rectum=4a=10
{\textbf{Step - 1: Find value of 'a' from Length of latus rectum,}}
{\text{We know that length of latus rectum of any parabola is 4a.}}
{\text{Given that 4a = 2 }} \Rightarrow {\text{a = }}\dfrac{1}{2}.
{\text{Also given that the axis is x + y = 2}}.
{\textbf{Step - 2: Use relation between Perpendicular distance from axis and Distance from tangent of a parabola.}}
{\text{From parabola's formula we know that,}}
\Rightarrow {\left( {{\text{perpendicular distance from axis}}} \right)^2}{\text{ = 4a}}\left( {{\text{distance from tangent}}} \right).
\Rightarrow {\text{ }}{\left( {\dfrac{{{\text{x + y}} - {\text{2}}}}{{\sqrt 2 }}} \right)^2} = 4\left( {\dfrac{1}{2}} \right)\left( {\dfrac{{{\text{x}} - {\text{y + 4}}}}{{\sqrt 2 }}} \right).
\Rightarrow {\left( {{\text{x + y}} - 2} \right)^2} = 2\sqrt 2 \left( {{\text{x}} - {\text{y + 4}}} \right).
{\textbf{Hence, The equation is }}{\left( {{\textbf{x + y}} - 2} \right)^2} = 2\sqrt 2 \left( {{\textbf{x}} - {\textbf{y + 4}}} \right).(C)
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