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Mathematics Question on Conic sections

The ends of the latus rectum of the conic x2+10x−16y+25=0x^ 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

(3 4), (-13, 4)

Explanation

Solution

The correct option is (C): (3 4), (- 13, 4)
Given that
x2+10x−16y+25=0x^{2}+10 x-16 y+25=0
⇒(x+5)2=16y\Rightarrow(x+5)^{2}=16 y
⇒X2=4AY\Rightarrow X^{2}=4 A Y
where X=x+5,A=4,Y=yX=x+5, A=4, Y=y
The ends of the latus rectum are (2 A, A) and (-2 A, A)
⇒x+5=2(4)⇒x=−8−5=3,y=4\Rightarrow x+5=2(4) \Rightarrow x=-8-5=3, y=4 and x+5=-2(4)
⇒x=−8−5=−13,y=4⇒(3,4)and(−13,4)\Rightarrow x=-8-5=-13, y=4 \Rightarrow(3,4) and (-13,4)