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10-2. Parabola, Ellipse, Hyperbola
normal
$P(6, 3)$ is a point on the hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$ . If the normal at point $P$ intersect the $x-$ axis at $(10, 0)$ , then the eccentricity of the hyperbola is
A
$\sqrt {\frac{5}{3}} $
B
$\frac{{\sqrt {13} }}{3}$
C
$\sqrt {\frac{5}{2}} $
D
$\frac{{\sqrt {13} }}{2}$
Solution
$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
Slope of normal is $-\frac{a^{2}}{2 b^{2}}$
$\therefore$ Equation of normal at $\mathrm{P}(6,3)$ is
$y-3=-\frac{a^{2}}{2 b^{2}}(x-6)$
at $(10,0) \Rightarrow-3=-\frac{a^{2}}{2 b^{2}}(10-6)$
$\Rightarrow \quad \frac{3}{2}=\frac{a^{2}}{b^{2}}$
$\therefore \quad e^{2}=1+\frac{b^{2}}{a^{2}}=1+\frac{2}{3}=\frac{5}{3}$
$\therefore \quad e=\sqrt{\frac{5}{3}}$
Standard 11
Mathematics
