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The equation of the directrix of the parabola `25{(x-2)^(2)+(y+5)^(2)}=(3x+4y-1)^(2),` isA. 3x+4y=0B. 3x+4y+1=0C. 4x-3y=0D. 3x+4y+1=0 |
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Answer» Correct Answer - B We have, `25{(x-2)^(2)+(y+5)^(2)}=(3x+4y-1)^(2),` `rArr" "sqrt((x-2)^(2)+(y+5)^(2))=|(3x+4y-1)/(sqrt(3^(2)+4^(2)))|` `rArr" SP=PM"`, where $ (2, 5) , P(x, y) and PM is the length of perpendicular from P on the line 3x+4y-1=0 Hence, 3x+4y-1=0 is the directrix adn $ (2, -5) is the focus of the given parabola. |
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