The number of normals that can be drawn from | vedclass

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(B) The equation of a normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with slope $m$ is given by:$y = mx - \frac{m(a^2 + b^2)}{\sqrt{a

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$4$

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(B) The equation of a normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with slope $m$ is given by:$y = mx - \frac{m(a^2 + b^2)}{\sqrt{a^2 - b^2m^2}}$Squaring both sides,we get:$(y - mx)^2 (a^2 - b^2m^2) = m^2(a^2 + b^2)^2$If this normal passes through a fixed point $(x_1, y_1)$,then:$(y_1 - mx_1)^2 (a^2 - b^2m^2) = m^2(a^2 + b^2)^2$Expanding this expression,we obtain a polynomial equation in $m$ of degree $4$:$(x_1^2b^2)m^4 - (2x_1y_1b^2)m^3 + (a^2b^2 + y_1^2b^2 - x_1^2a^2 + (a^2+b^2)^2)m^2 + (2x_1y_1a^2)m - y_1^2a^2 = 0$Since this is a quartic equation in $m$,there are at most $4$ real values of $m$ for any point $(x_1, y_1)$.Thus,the number of normals that can be drawn from an external point to the hyperbola is $4$.

The number of normals that can be drawn from an external point to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is:

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