Question
Mathematics Question on Conic sections
If a hyperbola passes through the point P(10,16) and it has vertices at (?6,0), then the equation of the normal to it at P is :
A
3x+4y=94
B
x+2y=42
C
2x+5y=100
D
x+3y=58
Answer
2x+5y=100
Explanation
Solution
36x2−b2y2...(i)
P(10,16) lies on (i) get b2=144
36x2−144y2=1
Equation of normal is
x1a2x+y1b2y=a2e2
2x+5y=100