JEE Advance - Mathematics (2011 - Paper 2 Offline - No. 7)
Let $$P(6, 3)$$ be a point on the hyperbola $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$. If the normal at the point $$P$$ intersects the $$x$$-axis at $$(9, 0)$$, then the eccentricity of the hyperbola is
$$\sqrt {{5 \over 2}} $$
$$\sqrt {{3 \over 2}} $$
$${\sqrt 2 }$$
$${\sqrt 3 }$$
คำอธิบาย
Equation of normal is
$$(y - 3) = {{ - {a^2}} \over {2{b^2}}}(x - 6) \Rightarrow {{{a^2}} \over {2{b^2}}} = 1 \Rightarrow e = \sqrt {{3 \over 2}} $$