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The set of all values of \mathrm{a^2} for which the line \mathrm{x+y=0} bisects two distinct chords drawn from a point \mathrm{\mathrm{P}\left(\frac{1+a}{2}, \frac{1-a}{2}\right)} on the circle \mathrm{2 x^2+2 y^2-(1+a) x-(1-a) y=0}, is equal to :

 

Option: 1

(0,4]
 


Option: 2

(4, \infty)
 


Option: 3

(2,12]
 


Option: 4

(8, \infty)


Answers (1)

best_answer

\mathrm{ x^2+y^2-\left(\frac{1+a}{2}\right) x-\left(\frac{1-a}{2}\right) y=0 }
\mathrm{ x\left(x-\left(\frac{1+a}{2}\right)\right)+y\left(y-\left(\frac{1-a}{2}\right)\right)=0 }

\mathrm{ y-y_1=m\left(x-x_1\right) \quad\left(x_1, y_1\right)=\left(\frac{1+a}{2}, \frac{1-a}{2}\right) }

\mathrm{ \& \: \mathrm{x}+\mathrm{y}=0 }

\mathrm{-\mathrm{x}-\mathrm{y}_1=\mathrm{mx}-\mathrm{mx}_1 }

\mathrm{ \mathrm{mx}_1-\mathrm{y}_1=(1+\mathrm{m}) \mathrm{x} }

\mathrm{x}=\frac{\mathrm{mx}_1-\mathrm{y}_1}{1+\mathrm{m}} \& \quad \mathrm{y}=\frac{\mathrm{y}_1-\mathrm{mx}_1}{1+\mathrm{m}}

\mathrm{ \frac{\frac{y_1}{2}-y}{\frac{x_1}{2}-x}=-\frac{1}{m} }

\mathrm{ \frac{\frac{y_1}{2}-\left(\frac{\mathrm{y}_1-m x_1}{1+m}\right)}{\frac{x_1}{2}-\left(\frac{m x_1-\mathrm{y}_1}{1+\mathrm{m}}\right)}=-\frac{1}{\mathrm{~m}} }

\mathrm{ =\frac{(1+\mathrm{m}) \mathrm{y}_1-2 \mathrm{y}_1+2 \mathrm{mx}_1}{(1+\mathrm{m}) \mathrm{x}_1-2 \mathrm{mx}_1+2 \mathrm{y}_1}=-\frac{1}{\mathrm{~m}} }

\mathrm{ \frac{\mathrm{m}\left(\mathrm{y}_1+2 \mathrm{x}_1\right)-\mathrm{y}_1}{-\mathrm{mx}_1+\mathrm{x}_1+2 \mathrm{y}_1}=-\frac{1}{\mathrm{~m}} }

\mathrm{\mathrm{~m}^2\left(\mathrm{y}_1+2 \mathrm{x}_1\right)-\mathrm{my}_1=\mathrm{mx}_1-\mathrm{x}_1-2 \mathrm{y}_1 }

\mathrm{ \mathrm{~m}^2\left(\mathrm{y}_1+2 \mathrm{x}_1\right)-\left(\mathrm{y}_1+\mathrm{x}_2\right) \mathrm{m}+\mathrm{x}_1+2 \mathrm{y}_1=0 }
\mathrm{ D>0 }
\mathrm{ \left(\mathrm{y}_1+\mathrm{x}_1\right)^2-4\left(\mathrm{y}_1+2 \mathrm{x}_1\right)\left(\mathrm{x}_2+2 \mathrm{y}_1\right)>0}

\mathrm{ \mathrm{x}_1=\frac{1+\mathrm{a}}{2}, \mathrm{y}_1=\frac{1-\mathrm{a}}{2} }

\mathrm{ \mathrm{x}_1+\mathrm{y}_1=1 }

\mathrm{ \mathrm{y}_1+2 \mathrm{x}_1=\frac{1-\mathrm{a}}{2}+1+\mathrm{a}=\frac{3}{2}-\frac{\mathrm{a}}{2}=\frac{3-\mathrm{a}}{2} }

\mathrm{ \mathrm{x}_1+2 \mathrm{y}_1=\frac{1+\mathrm{a}}{2}+1-\mathrm{a}=\frac{3}{2}+\frac{\mathrm{a}}{2}=\frac{3+\mathrm{a}}{2} }

\mathrm{ 1-4\left(\frac{3-\mathrm{a}}{2}\right)\left(\frac{3+\mathrm{a}}{2}\right)>0 }

\mathrm{ 1-\left(9-\mathrm{a}^2\right)>0 }
\mathrm{ \mathrm{a}^2-8>0 \quad \rightarrow(8, \infty) }

\mathrm{ \mathrm{a}^2>8 }
Ans. 4









 

Posted by

avinash.dongre

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