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Question

Mathematics Question on Conic sections

If a hyperbola passes through the point P(10,16)P(10,16) and it has vertices at (?6,0)( ? 6,0), then the equation of the normal to it at P is :

A

3x+4y=943x + 4y=94

B

x+2y=42x + 2y=42

C

2x+5y=1002x + 5y=100

D

x+3y=58x + 3y=58

Answer

2x+5y=1002x + 5y=100

Explanation

Solution

x236y2b2...(i)\frac{x^{2}}{36} - \frac{y^{2}}{b^{2}} \quad...\left(i\right)
P(10,16)P\left(10,16\right) lies on (i)\left(i\right) get b2=144b^{2} = 144
x236y2144=1\frac{x^{2}}{36} - \frac{y^{2}}{144} = 1
Equation of normal is
a2xx1+b2yy1=a2e2\frac{a^{2}x}{x_{1}} + \frac{b^{2}y}{y_{1}} = a^{2}e^{2}
2x+5y=1002x + 5y = 100