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Grade: 11
        
The equation of one of the latusrectum of the hyperbola (10x - 5)2+ (10y - 2)2 = 9(3x + 4y - 7)2 is ax + by + c = 0, then the value of |b + 2c| is
one year ago

Answers : (1)

Arun
22985 Points
							
 it can be written (x−1/2)²+(y−1/5)² = (9/4)|(3x+4y−7)/5|² 

→ √{(x−1/2)²+(y−1/5)²} = (3/2)| (3x+4y−7)/5 | … (ii) 

LHS is distance from (1/2,1/5) and RHS is 3/2 X distance from line 3x+4y−7=0 

So (ii) is focus-directrix definition of hyperbola with e=3/2 
(1/2,1/5) is a focus and 3x+4y−7=0 is a directrix 

The latus rectum through this focus is || to the directrix 
Its equation is thus 3(x−1/2) + 4(y−1/5) = 0 → 30x+40y−23=0 


To find other latus rectum first find centre by differentiating (i) wrt x & y and solving 

dwrt x : 20(10x−5) = 54(3x+4y−7) 

dwrt y : 20(10y−2) = 72(3x+4y−7) 

Solution is x = 947/625, y = 971/625 


Second focus is thus ( 1/2 + 2(947/625−1/2), 1/5 + 2(971/625−1/5) ) 

= ( 3163/1250, 3634/1250 ) 


Second latus rectum has equation 3(x−3163/1250) + 4(y−3634/1250) = 0 

→ 150x+200y−961
one year ago
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