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10-2. Parabola, Ellipse, Hyperbola
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The equation of the normal at the point $(6, 4)$ on the hyperbola $\frac{{{x^2}}}{9} - \frac{{{y^2}}}{{16}} = 3$, is
A
$3x + 8y = 50$
B
$3x - 8y = 50$
C
$8x + 3y = 50$
D
$8x - 3y = 50$
Solution
(a) Equation of normal at any point $({x_1},{y_1})$ on hyperbola is,
$\frac{{{a^2}(x – {x_1})}}{{{x_1}}} = \frac{{{b^2}(y – {y_1})}}{{ – {y_1}}}$
Here, ${a^2} = 267,\,{b^2} = 48$ and $({x_1},{y_1}) = (6,4)$
$\therefore \frac{{27(x – 6)}}{6} = – \frac{{48(y – 4)}}{4}$
==> $3(x – 6) = – 8(y – 4)$
==> $3x + 8y = 50$.
Standard 11
Mathematics
