JEE MAIN - Mathematics (2019 - 9th January Evening Slot - No. 22)
A hyperbola has its centre at the origin, passes through the point (4, 2) and has transverse axis of length 4 along the x-axis. Then the eccentricity of the hyperbola is :
$${3 \over 2}$$
$$\sqrt 3 $$
2
$${2 \over {\sqrt 3 }}$$
Explanation
Let the equation of hyperbola
$${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}}$$ = 1
Given 2a = 4
$$ \Rightarrow $$ $$a$$ = 2
It passes through (4, 2)
$$ \therefore $$ $${{16} \over 4} - {4 \over {{b^2}}}$$ = 1
$$ \Rightarrow $$ b2 = $${4 \over 3}$$
e = $$\sqrt {1 + {{{b^2}} \over {{a^2}}}} $$ = $$\sqrt {1 + {{4/3} \over 4}} $$
= $$\sqrt {1 + {1 \over 3}} $$ = $${2 \over {\sqrt 3 }}$$
$${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}}$$ = 1
Given 2a = 4
$$ \Rightarrow $$ $$a$$ = 2
It passes through (4, 2)
$$ \therefore $$ $${{16} \over 4} - {4 \over {{b^2}}}$$ = 1
$$ \Rightarrow $$ b2 = $${4 \over 3}$$
e = $$\sqrt {1 + {{{b^2}} \over {{a^2}}}} $$ = $$\sqrt {1 + {{4/3} \over 4}} $$
= $$\sqrt {1 + {1 \over 3}} $$ = $${2 \over {\sqrt 3 }}$$
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