JEE MAIN - Mathematics (2016 - 10th April Morning Slot - No. 15)
A hyperbola whose transverse axis is along the major axis of the conic, $${{{x^2}} \over 3} + {{{y^2}} \over 4} = 4$$ and has vertices at the foci of this conic. If the eccentricity of the hyperbola is $${3 \over 2},$$ then which of the following points does NOT lie on it?
(0, 2)
$$\left( {\sqrt 5 ,2\sqrt 2 } \right)$$
$$\left( {\sqrt {10} ,2\sqrt 3 } \right)$$
$$\left( {5,2\sqrt 3 } \right)$$
Explanation
$${{{x^2}} \over {12}} + {{{y^2}} \over {16}}$$ = 1
e = $$\sqrt {1 - {{12} \over {16}}} $$ = $${1 \over 2}$$
Foci (0, 2) & (0, $$-$$ 2)
So, transverse axis of hyperbola
= 2b = 4
$$ \Rightarrow $$ b = 2 & a2 = 12 (e2 $$-$$ 1)
$$ \Rightarrow $$ a2 = 4$$\left( {{9 \over 4} - 1} \right)$$
$$ \Rightarrow $$ a2 = 5
$$ \therefore $$ It's equation is $${{{x^2}} \over 5} - {{{y^2}} \over 4}$$ = $$-$$ 1
The point (5, 2$$\sqrt 3 $$) does not satisfy the above equation.
e = $$\sqrt {1 - {{12} \over {16}}} $$ = $${1 \over 2}$$
Foci (0, 2) & (0, $$-$$ 2)
So, transverse axis of hyperbola
= 2b = 4
$$ \Rightarrow $$ b = 2 & a2 = 12 (e2 $$-$$ 1)
$$ \Rightarrow $$ a2 = 4$$\left( {{9 \over 4} - 1} \right)$$
$$ \Rightarrow $$ a2 = 5
$$ \therefore $$ It's equation is $${{{x^2}} \over 5} - {{{y^2}} \over 4}$$ = $$-$$ 1
The point (5, 2$$\sqrt 3 $$) does not satisfy the above equation.
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