JEE MAIN - Mathematics (2019 - 10th April Morning Slot - No. 19)
If a directrix of a hyperbola centred at the origin and passing through the point (4, –2$$\sqrt 3 $$ ) is 5x = 4$$\sqrt 5 $$ and
its eccentricity is e, then :
4e4 – 24e2 + 27 = 0
4e4 – 24e2 + 35 = 0
4e4 – 12e2 - 27 = 0
4e4 + 8e2 - 35 = 0
Explanation
5x = 4$$\sqrt 5 $$
$$ \Rightarrow $$ $$x = {4 \over {\sqrt 5 }}$$
_10th_April_Morning_Slot_en_19_2.png)
Equation of hyperbola
$${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$ it passes through $$\left( {4, - 2\sqrt 3 } \right)$$
$$ \therefore {e^2} = 1 + {{{b^2}} \over {{a^2}}}$$
$$ \Rightarrow {a^2}{e^2} - {a^2} = {b^2}$$
$$ \Rightarrow {{16} \over {{a^2}}} - {{12} \over {{a^2}{e^2} - {a^2}}} = 1$$
$$ \Rightarrow {4 \over {{a^2}}}\left[ {{4 \over 1} - {3 \over {{e^2} - 1}}} \right] = 1$$
$$ \Rightarrow 4{e^2} - 4 - 3 = ({e^2} - 1)\left( {{{{a^2}} \over 4}} \right)$$
$$ \Rightarrow 4(4{e^3} - 7) = ({e^2} - 1){\left( {{{4e} \over {\sqrt 5 }}} \right)^2}$$
$$ \Rightarrow 4{e^4} - 24{e^2} + 35 = 0$$
$$ \Rightarrow $$ $$x = {4 \over {\sqrt 5 }}$$
Equation of hyperbola
$${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$ it passes through $$\left( {4, - 2\sqrt 3 } \right)$$
$$ \therefore {e^2} = 1 + {{{b^2}} \over {{a^2}}}$$
$$ \Rightarrow {a^2}{e^2} - {a^2} = {b^2}$$
$$ \Rightarrow {{16} \over {{a^2}}} - {{12} \over {{a^2}{e^2} - {a^2}}} = 1$$
$$ \Rightarrow {4 \over {{a^2}}}\left[ {{4 \over 1} - {3 \over {{e^2} - 1}}} \right] = 1$$
$$ \Rightarrow 4{e^2} - 4 - 3 = ({e^2} - 1)\left( {{{{a^2}} \over 4}} \right)$$
$$ \Rightarrow 4(4{e^3} - 7) = ({e^2} - 1){\left( {{{4e} \over {\sqrt 5 }}} \right)^2}$$
$$ \Rightarrow 4{e^4} - 24{e^2} + 35 = 0$$
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