JEE MAIN - Mathematics (2016 - 9th April Morning Slot - No. 12)
Let a and b respectively be the semitransverse and semi-conjugate axes of a
hyperbola whose eccentricity satisfies the equation 9e2 − 18e + 5 = 0. If S(5, 0) is a focus and 5x = 9 is the corresponding directrix of this hyperbola, then a2 − b2 is equal to :
7
$$-$$ 7
5
$$-$$ 5
Explanation
As S(5, 0) is the focus.
$$ \therefore $$ ae = 5 . . . (1)
As 5x = 9
$$ \therefore $$ x = $${9 \over 5}$$ is the directrix
As we know directrix = $${a \over e}$$
$$ \therefore $$ $${a \over e} = {9 \over 5}$$ . . . .(2)
Solving (1) and (2), we get
a = 3 and e = $${5 \over 3}$$
As we know,
b2 = a2 (e2 $$-$$ 1) = 9$$\left( {{{25} \over 9} - 1} \right)$$ = 16
$$ \therefore $$ a2 $$-$$ b2 = 9 $$-$$ 16 $$=$$ $$-$$ 7
$$ \therefore $$ ae = 5 . . . (1)
As 5x = 9
$$ \therefore $$ x = $${9 \over 5}$$ is the directrix
As we know directrix = $${a \over e}$$
$$ \therefore $$ $${a \over e} = {9 \over 5}$$ . . . .(2)
Solving (1) and (2), we get
a = 3 and e = $${5 \over 3}$$
As we know,
b2 = a2 (e2 $$-$$ 1) = 9$$\left( {{{25} \over 9} - 1} \right)$$ = 16
$$ \therefore $$ a2 $$-$$ b2 = 9 $$-$$ 16 $$=$$ $$-$$ 7
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