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Find the locus of the circumcentre of the variable triangle having x-axis, \mathrm{x=2} and \mathrm{m x+l y=1} as its sides where \mathrm{(m, l)} lies on the parabola \mathrm{ x^2=4} by.

Option: 1

\mathrm{y=8 b\left(x^2-3 x+2\right)}


Option: 2

\mathrm{y=4 b\left(x^2+3 x-2\right)}


Option: 3

\mathrm{y=8 b\left(x^2+3 x-2\right)}


Option: 4

\mathrm{y=4 b\left(x^2+3 x+2\right)}


Answers (1)

best_answer

Here triangle formed by x-axis,\mathrm{ x=2} and \mathrm{ m x+l y=1} is a right angled triangle having \mathrm{ m x+l y=1} as its hypotenuse. Circumcentre of the triangle will be mid-point of hypotenuse. Let circumcentre of the triangle is \mathrm{ (h, k)}. Here extremities of hypotenuse are point of intersection of hypotenuse with x-axis and with \mathrm{ \mathrm{x}=2} respectively.
Therefore \mathrm{h=\frac{\frac{1}{m}+2}{2}} and \mathrm{k=\frac{1-2 m}{2 l}}

Or \mathrm{ 2 \mathrm{~h}-2=\frac{1}{\mathrm{~m}} \text { and } 2 \mathrm{k}=\frac{1-2 \mathrm{~m}}{\mathrm{l}} }

Also \mathrm{\mathrm{m}^2=4 \mathrm{bl} (since \, \, \mathrm{m}, \, \, I \, \, lies \, \, on \, \, the \, \, parabola \, \, \mathrm{x}^2=4 \mathrm{by} )}

From (1) and (2)

\mathrm{ \frac{k}{2 b}=4\left(h^2-3 h+2\right) \text { or } y=8 b\left(x^2-3 x+2\right) }

Posted by

jitender.kumar

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