The equation of the normal at the point (6, 4 | vedclass

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Question

(a) Equation of normal at any point $({x_1},{y_1})$ on hyperbola is,

$\frac{{{a^2}(x - {x_1})}}{{{x_1}}} = \frac{{{b^2}(y - {y_1})}}{{ - {y_1}}}$

Vedclass · Accepted Answer

$3x + 8y = 50$

Vedclass · Answer

(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 $({x_1},{y_1}) = (6,4)$

$	herefore \frac{{27(x - 6)}}{6} = - \frac{{48(y - 4)}}{4}$

==> $3(x - 6) = - 8(y - 4)$

==> $3x + 8y = 50$.

The equation of the normal at the point $(6, 4)$ on the hyperbola $\frac{{{x^2}}}{9} - \frac{{{y^2}}}{{16}} = 3$, is

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