Answer on Question #56310 - Math - Analytic Geometry
Find the equation of the pair of straight lines joining the origin to the points of intersection of the circles represented by x 2 + y 2 = a 2 x^{2} + y^{2} = a^{2} x 2 + y 2 = a 2 x 2 + y 2 + 2 ( g x + f y ) = 0 x^{2} + y^{2} + 2(gx + fy) = 0 x 2 + y 2 + 2 ( gx + f y ) = 0
Solution
Let's find points of intersection of these circles
{ x 2 + y 2 = a 2 x 2 + y 2 + 2 ( g x + f y ) = 0 \left\{ \begin{array}{c} x ^ {2} + y ^ {2} = a ^ {2} \\ x ^ {2} + y ^ {2} + 2 (g x + f y) = 0 \end{array} \right. { x 2 + y 2 = a 2 x 2 + y 2 + 2 ( gx + f y ) = 0 { x 2 + y 2 = a 2 a 2 + 2 ( g x + f y ) = 0 \left\{ \begin{array}{c} x ^ {2} + y ^ {2} = a ^ {2} \\ a ^ {2} + 2 (g x + f y) = 0 \end{array} \right. { x 2 + y 2 = a 2 a 2 + 2 ( gx + f y ) = 0 { x 2 + y 2 = a 2 x = − a 2 2 g − f g y \left\{ \begin{array}{l} x ^ {2} + y ^ {2} = a ^ {2} \\ x = - \frac {a ^ {2}}{2 g} - \frac {f}{g} y \end{array} \right. { x 2 + y 2 = a 2 x = − 2 g a 2 − g f y ( a 2 2 g + f g y ) 2 + y 2 = a 2 \left(\frac {a ^ {2}}{2 g} + \frac {f}{g} y\right) ^ {2} + y ^ {2} = a ^ {2} ( 2 g a 2 + g f y ) 2 + y 2 = a 2 ( f 2 g 2 + 1 ) y 2 + a 2 f g 2 y + a 2 ( a 2 4 g 2 − 1 ) = 0 \left(\frac {f ^ {2}}{g ^ {2}} + 1\right) y ^ {2} + \frac {a ^ {2} f}{g ^ {2}} y + a ^ {2} \left(\frac {a ^ {2}}{4 g ^ {2}} - 1\right) = 0 ( g 2 f 2 + 1 ) y 2 + g 2 a 2 f y + a 2 ( 4 g 2 a 2 − 1 ) = 0 y = − a 2 f g 2 ± 4 a 2 + 4 f 2 a 2 g 2 − a 4 g 2 2 ( f 2 g 2 + 1 ) y = \frac {- \frac {a ^ {2} f}{g ^ {2}} \pm \sqrt {4 a ^ {2} + 4 \frac {f ^ {2} a ^ {2}}{g ^ {2}} - \frac {a ^ {4}}{g ^ {2}}}}{2 \left(\frac {f ^ {2}}{g ^ {2}} + 1\right)} y = 2 ( g 2 f 2 + 1 ) − g 2 a 2 f ± 4 a 2 + 4 g 2 f 2 a 2 − g 2 a 4
Thus
{ x = − a 2 g ∓ 4 a 2 + 4 f 2 a 2 g 2 − a 4 g 2 2 ( f 2 g 2 + 1 ) y = − a 2 f g 2 ± 4 a 2 + 4 f 2 a 2 g 2 − a 4 g 2 2 ( f 2 g 2 + 1 ) \left\{ \begin{array}{l} x = \frac {- \frac {a ^ {2}}{g} \mp \sqrt {4 a ^ {2} + 4 \frac {f ^ {2} a ^ {2}}{g ^ {2}} - \frac {a ^ {4}}{g ^ {2}}}}{2 \left(\frac {f ^ {2}}{g ^ {2}} + 1\right)} \\ y = \frac {- \frac {a ^ {2} f}{g ^ {2}} \pm \sqrt {4 a ^ {2} + 4 \frac {f ^ {2} a ^ {2}}{g ^ {2}} - \frac {a ^ {4}}{g ^ {2}}}}{2 \left(\frac {f ^ {2}}{g ^ {2}} + 1\right)} \end{array} \right. ⎩ ⎨ ⎧ x = 2 ( g 2 f 2 + 1 ) − g a 2 ∓ 4 a 2 + 4 g 2 f 2 a 2 − g 2 a 4 y = 2 ( g 2 f 2 + 1 ) − g 2 a 2 f ± 4 a 2 + 4 g 2 f 2 a 2 − g 2 a 4
The equation of the line pathing through point
( − a 2 g − 4 a 2 + 4 f 2 a 2 g 2 − a 4 g 2 2 ( f 2 g 2 + 1 ) , − a 2 f g 2 + 4 a 2 + 4 f 2 a 2 g 2 − a 4 g 2 2 ( f 2 g 2 + 1 ) ) \left(\frac {- \frac {a ^ {2}}{g} - \sqrt {4 a ^ {2} + 4 \frac {f ^ {2} a ^ {2}}{g ^ {2}} - \frac {a ^ {4}}{g ^ {2}}}}{2 \left(\frac {f ^ {2}}{g ^ {2}} + 1\right)}, \frac {- \frac {a ^ {2} f}{g ^ {2}} + \sqrt {4 a ^ {2} + 4 \frac {f ^ {2} a ^ {2}}{g ^ {2}} - \frac {a ^ {4}}{g ^ {2}}}}{2 \left(\frac {f ^ {2}}{g ^ {2}} + 1\right)}\right) ⎝ ⎛ 2 ( g 2 f 2 + 1 ) − g a 2 − 4 a 2 + 4 g 2 f 2 a 2 − g 2 a 4 , 2 ( g 2 f 2 + 1 ) − g 2 a 2 f + 4 a 2 + 4 g 2 f 2 a 2 − g 2 a 4 ⎠ ⎞
is
y = a 2 f g 2 − 4 a 2 + 4 f 2 a 2 g 2 − a 4 g 2 a 2 g + 4 a 2 + 4 f 2 a 2 g 2 − a 4 g 2 x y = \frac {\frac {a ^ {2} f}{g ^ {2}} - \sqrt {4 a ^ {2} + 4 \frac {f ^ {2} a ^ {2}}{g ^ {2}} - \frac {a ^ {4}}{g ^ {2}}}}{\frac {a ^ {2}}{g} + \sqrt {4 a ^ {2} + 4 \frac {f ^ {2} a ^ {2}}{g ^ {2}} - \frac {a ^ {4}}{g ^ {2}}}} x y = g a 2 + 4 a 2 + 4 g 2 f 2 a 2 − g 2 a 4 g 2 a 2 f − 4 a 2 + 4 g 2 f 2 a 2 − g 2 a 4 x
And the line pathing through point
( − a 2 g + 4 a 2 + 4 f 2 a 2 g 2 − a 4 g 2 2 ( f 2 g 2 + 1 ) , − a 2 f g 2 − 4 a 2 + 4 f 2 a 2 g 2 − a 4 g 2 2 ( f 2 g 2 + 1 ) ) \left(\frac {- \frac {a ^ {2}}{g} + \sqrt {4 a ^ {2} + 4 \frac {f ^ {2} a ^ {2}}{g ^ {2}} - \frac {a ^ {4}}{g ^ {2}}}}{2 \left(\frac {f ^ {2}}{g ^ {2}} + 1\right)}, \frac {- \frac {a ^ {2} f}{g ^ {2}} - \sqrt {4 a ^ {2} + 4 \frac {f ^ {2} a ^ {2}}{g ^ {2}} - \frac {a ^ {4}}{g ^ {2}}}}{2 \left(\frac {f ^ {2}}{g ^ {2}} + 1\right)}\right) ⎝ ⎛ 2 ( g 2 f 2 + 1 ) − g a 2 + 4 a 2 + 4 g 2 f 2 a 2 − g 2 a 4 , 2 ( g 2 f 2 + 1 ) − g 2 a 2 f − 4 a 2 + 4 g 2 f 2 a 2 − g 2 a 4 ⎠ ⎞
is
y = a 2 f g 2 + 4 a 2 + 4 f 2 a 2 g 2 − a 4 g 2 a 2 g − 4 a 2 + 4 f 2 a 2 g 2 − a 4 g 2 x y = \frac {\frac {a ^ {2} f}{g ^ {2}} + \sqrt {4 a ^ {2} + 4 \frac {f ^ {2} a ^ {2}}{g ^ {2}} - \frac {a ^ {4}}{g ^ {2}}}}{\frac {a ^ {2}}{g} - \sqrt {4 a ^ {2} + 4 \frac {f ^ {2} a ^ {2}}{g ^ {2}} - \frac {a ^ {4}}{g ^ {2}}}} x y = g a 2 − 4 a 2 + 4 g 2 f 2 a 2 − g 2 a 4 g 2 a 2 f + 4 a 2 + 4 g 2 f 2 a 2 − g 2 a 4 x
Answer: y = a 2 f g 2 − 4 a 2 + 4 f 2 a 2 g 2 − a 4 g 2 a 2 g + 4 a 2 + 4 f 2 a 2 g 2 − a 4 g 2 x y = \frac{\frac{a^2f}{g^2} - \sqrt{4a^2 + 4\frac{f^2a^2}{g^2} - \frac{a^4}{g^2}}}{\frac{a^2}{g} + \sqrt{4a^2 + 4\frac{f^2a^2}{g^2} - \frac{a^4}{g^2}}} x y = g a 2 + 4 a 2 + 4 g 2 f 2 a 2 − g 2 a 4 g 2 a 2 f − 4 a 2 + 4 g 2 f 2 a 2 − g 2 a 4 x
and
y = a 2 f g 2 + 4 a 2 + 4 f 2 a 2 g 2 − a 4 g 2 a 2 g − 4 a 2 + 4 f 2 a 2 g 2 − a 4 g 2 x y = \frac {\frac {a ^ {2} f}{g ^ {2}} + \sqrt {4 a ^ {2} + 4 \frac {f ^ {2} a ^ {2}}{g ^ {2}} - \frac {a ^ {4}}{g ^ {2}}}}{\frac {a ^ {2}}{g} - \sqrt {4 a ^ {2} + 4 \frac {f ^ {2} a ^ {2}}{g ^ {2}} - \frac {a ^ {4}}{g ^ {2}}}} x y = g a 2 − 4 a 2 + 4 g 2 f 2 a 2 − g 2 a 4 g 2 a 2 f + 4 a 2 + 4 g 2 f 2 a 2 − g 2 a 4 x
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#340153 on Dec 2023