The length of the latus rectum of the parabola
13[(x-3)^2+(y-4)^2 )= (2x-3y+ 5)^2 is
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The length of the latus rectum of the parabola
13[(x-3)^2+(y-4)^2 )= (2x-3y+ 5)^2 is
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SOLUTION :
TO DETERMINE
The length of the latus rectum of the parabola
[tex] \sf{ 13 \bigg[ {(x - 3)}^{2} + {(y - 4)}^{2}\bigg] = {(2x - 3y + 5)}^{2} \: }[/tex]
EVALUATION
Here the given equation of parabola is
[tex] \sf{ 13 \bigg[ {(x - 3)}^{2} + {(y - 4)}^{2}\bigg] = {(2x - 3y + 5)}^{2} \: }[/tex]
So the focus of the parabola is ( 3 , 4 )
Also the equation of the directrix is
[tex] \sf{2x - 3y + 5 = 0 \: }[/tex]
Now the distance between focus and directrix is
[tex] \displaystyle\sf{ = \bigg| \frac{(2 \times 3) - (3 \times 4) + 5}{ \sqrt{ {2}^{2} + {3}^{2} } } \bigg| }[/tex]
[tex] \displaystyle\sf{ = \bigg| \frac{6 - 12 + 5}{ \sqrt{ 4 + 9 } } \bigg| }[/tex]
[tex] \displaystyle\sf{ = \frac{1}{ \sqrt{ 13 } } }[/tex]
Now we know that Latus rectum is double of the distance between focus and directrix
Hence the length of the latus rectum
[tex] \displaystyle\sf{ = 2 \times \frac{1}{ \sqrt{ 13 } } } \: \: \: unit[/tex]
[tex] \displaystyle\sf{ = \frac{2}{ \sqrt{ 13 } } } \: \: \: unit[/tex]
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