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  • The tangent and normal at a point to the parabola <img alt="\ma

The tangent and normal at a point \mathrm{P \equiv(16,16)} to the parabola \mathrm{y^2=16 x} intersect the x-axis at A and B respectively. Circle \mathrm{C_1} is drawn through \mathrm{P, A} and B. Another circle \mathrm{C_2 x^2+y^2+4 x +2 y+10=0} cuts the parabola orthogonally and passes through P. Find the equation of the common chord of \mathrm{\mathrm{C}_1} and \mathrm{\mathrm{C}_2.}

Option: 1

\mathrm{x-7 y+197=0}


Option: 2

\mathrm{x-4 y+199=0}


Option: 3

\mathrm{x-7 y+291=0}


Option: 4

\mathrm{\text { None of these }}


Answers (1)

best_answer

By property; Centre of circle \mathrm{ \mathrm{C}_1 \equiv} focus of parabola

\mathrm{ \begin{aligned} & \equiv(4,0) \\ & \Rightarrow \text { Equation of circle } \mathrm{C}_1 \\ & \therefore(\mathrm{x}-4)^2+\mathrm{y}^2=(16-4)^2+16^2 \\ & \Rightarrow \mathrm{x}^2+\mathrm{y}^2-8 \mathrm{x}-384=0 \end{aligned} }

Since parabola and circle \mathrm{\mathrm{C}_2} cut each other orthogonally

centre of \mathrm{\mathrm{C}_2 \equiv(-2,-\mathrm{f})} lies on the tangent AP.

Equation of \mathrm{A P} is \mathrm{16 y=8(x+16) \Rightarrow-2 f=14 \Rightarrow f=-7}

\mathrm{\Rightarrow} Equation of circle \mathrm{\mathrm{C}_2} is \mathrm{\mathrm{x}^2+\mathrm{y}^2+4 \mathrm{x}-14 \mathrm{y}+10=0}

\mathrm{\Rightarrow} Equation of common chord \mathrm{MP =\mathrm{C}_1-\mathrm{C}_2=0}

\mathrm{\Rightarrow 12 x-14 y+394=0 \Rightarrow x-7 y+197=0.}

Posted by

Irshad Anwar

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