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 :

Question

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

Tardigrade Admin · Accepted Answer

(x2/36) - (y2/b2) ...(i) P(10,16) lies on (i) get b2 = 144 (x2/36) - (y2/144) = 1 Equation of normal is (a2x/x1) + (b2y/y1) = a2e2 2x + 5y = 100

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

$3 x + 4 y = 94$

$x + 2 y = 42$

$2 x + 5 y = 100$

$x + 3 y = 58$

$\frac{x ^{2}}{36} - \frac{y ^{2}}{b ^{2}} ... (i)$
$P (10, 16)$ lies on $(i)$ get $b^{2} = 144$
$\frac{x ^{2}}{36} - \frac{y ^{2}}{144} = 1$
Equation of normal is
$\frac{a ^{2} x}{x _{1}} + \frac{b ^{2} y}{y _{1}} = a^{2} e^{2}$
$2 x + 5 y = 100$

Q. 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 :

3x+4y=94

x+2y=42

2x+5y=100

x+3y=58