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  • The circle <img alt="\mathrm{x^{2}+y^{2}+2 g x+m 2 f y+c=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E%7B2%7D&plus;y%5E%7B2%7D&plus;2%20g%20x&plus;m%202%20f%

The circle \mathrm{x^{2}+y^{2}+2 g x+m 2 f y+c=0}, cuts the parabola \mathrm{x^{2}=4ay}at points \mathrm{A_{i}\left(x_{i}, y_{i}\right), i=1,2,3,4}; then 

Option: 1

\mathrm{ \sum y_{i}=0}


Option: 2

\mathrm{\sum y_{i}=-4(f+2 a)}


Option: 3

\mathrm{\sum x_{i}=-4(g+2 a)}


Option: 4

\mathrm{\sum x_{i}=-2(g+2 a)}\mathrm{\sum x_{i}=-2(g+2 a)}


Answers (1)

best_answer

Putting \mathrm{y=\frac{x^{2}}{4 a}} in the equation of circle, we get
\mathrm{ x^{2}+\frac{x^{4}}{16 a^{2}}+2 g x+\frac{f x^{2}}{2 a}+c=0}

\mathrm{\Rightarrow \quad x^{4}+x^{2}\left(16 a^{2}+8 a f\right)+32 a^{2} g x+16 a^{2} c=0 }


\mathrm{\Rightarrow \quad \sum x_{1} x_{2}=8 a(f+2 a) },
\mathrm{ \sum x_{1} x_{2} x_{3}=-32 a^{2} g, \sum x_{1} x_{2} x_{3} x_{4}=16 a^{2} c }
\mathrm{ \Rightarrow \quad \sum y_{i}=\frac{1}{4 a} \sum x_{1}^{2}=\frac{1}{4 a}\left[\left(\sum x_{1}\right)^{2}-2 \sum x_{1} x_{2}\right] }
\mathrm{ =-4(f+2 a)}

Posted by

Kshitij

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