Question

The ends of the latus rectum of the conic \(x^ 2 + 10x - 16y + 25 = 0\) are

• A. (3, -4), (13, 4)
• B. ( -3 ,-4 ) , (13 ,-4 )
• C. (3 4), (-13, 4)
• D. (5, -8), (-5, 8)

Answer 1

The Correct Option is C

The correct option is (C): (3 4), (- 13, 4)
Given that 
\(x^{2}+10 x-16 y+25=0\)
\(\Rightarrow(x+5)^{2}=16 y\)
\(\Rightarrow X^{2}=4 A Y\)
where \(X=x+5, A=4, Y=y\)
The ends of the latus rectum are (2 A, A) and (-2 A, A) 
\(\Rightarrow x+5=2(4) \Rightarrow x=-8-5=3, y=4\) and x+5=-2(4) 
\(\Rightarrow x=-8-5=-13, y=4 \Rightarrow(3,4) and (-13,4)\)