JEE MAIN - Mathematics (2019 - 10th April Evening Slot - No. 19)
If 5x + 9 = 0 is the directrix of the hyperbola 16x2
– 9y2
= 144, then its corresponding focus is :
$$\left( {{5 \over 3},0} \right)$$
(5, 0)
(- 5, 0)
$$\left( { - {5 \over 3},0} \right)$$
Explanation
$${{{x^2}} \over 9} - {{{y^2}} \over {16}} = 1$$
$$ \therefore $$ a = 3 and b = 4
$${e^2} = 1 + {{{b^2}} \over {{a^2}}}$$
$$ \Rightarrow {e^2} = 1 + {{16} \over 9}$$
$$ \Rightarrow $$ e = $$5 \over 3$$
$$ \therefore $$ focus is (–ae, 0) = (–5, 0)
$$ \therefore $$ a = 3 and b = 4
$${e^2} = 1 + {{{b^2}} \over {{a^2}}}$$
$$ \Rightarrow {e^2} = 1 + {{16} \over 9}$$
$$ \Rightarrow $$ e = $$5 \over 3$$
$$ \therefore $$ focus is (–ae, 0) = (–5, 0)
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