JEE MAIN - Mathematics (2019 - 9th January Morning Slot - No. 7)
Let $$0 < \theta < {\pi \over 2}$$. If the eccentricity of the
hyperbola $${{{x^2}} \over {{{\cos }^2}\theta }} - {{{y^2}} \over {{{\sin }^2}\theta }}$$ = 1 is greater
than 2, then the length of its latus rectum lies in the interval :
hyperbola $${{{x^2}} \over {{{\cos }^2}\theta }} - {{{y^2}} \over {{{\sin }^2}\theta }}$$ = 1 is greater
than 2, then the length of its latus rectum lies in the interval :
(3, $$\infty $$)
$$\left( {{3 \over 2},2} \right]$$
$$\left( {1,{3 \over 2}} \right]$$
$$\left( {2,3} \right]$$
Explanation
Given hyperbola,
$${{{x^2}} \over {{{\cos }^2}\theta }} - {{{y^2}} \over {{{\sin }^2}\theta }} = 1$$
here a = cos$$\theta $$
and b = sin$$\theta $$
We know, eccentricity of the hyperbola is,
$$\sqrt {1 + {{{b^2}} \over {{a^2}}}} $$
$$ \therefore $$ Here eccentricity
(e) = $$\sqrt {1 + {{{{\sin }^2}\theta } \over {{{\cos }^2}\theta }}} $$
Given that,
$$\sqrt {1 + {{{{\sin }^2}\theta } \over {{{\cos }^2}\theta }}} > 2$$
$$ \Rightarrow $$ $$\sqrt {1 + {{\tan }^2}\theta } > 2$$
$$ \Rightarrow $$ 1 + tan2$$\theta $$ > 4
$$ \Rightarrow $$ tan2$$\theta $$ > 3
$$ \Rightarrow $$ tan$$\theta $$ > $$ \pm \sqrt 3 $$
As given $$\theta $$ $$ \in $$ $$\left( {0,{\pi \over 2}} \right)$$
possible value of tan$$\theta $$ > $$\sqrt 3 $$
So, $$\theta $$ can be in the range $${\pi \over 3} < \theta < {\pi \over 2}$$
_9th_January_Morning_Slot_en_7_1.png)
We know latus ractum (LR) = $${{2{b^2}} \over a}$$
$$ \therefore $$ LR = $${{2{{\sin }^2}\theta } \over {\cos \theta }}$$
= 2 tan$$\theta $$ sin$$\theta $$
We know in the range $${\pi \over 3} < \theta < {\pi \over 2}$$ tan$$\theta $$ and sin$$\theta $$ both are increasing function.
So, at $${\pi \over 3}$$ value of LR will be minimum and at $${\pi \over 2}$$ value of LR will be maximum.
$$ \therefore $$ Minimum value of LR = 2tan$${\pi \over 3}$$ sin$${\pi \over 3}$$
= 2 $$ \times $$ $$\sqrt 3 \times {{\sqrt 3 } \over 2}$$
= 3
Maximum value of LR = 2tan$${\pi \over 2}$$ sin$${\pi \over 2}$$
= 2$$\left( \infty \right) \times 1$$
= $$\infty $$
$$ \therefore $$ Interval of LR = (3, $$\infty $$)
$${{{x^2}} \over {{{\cos }^2}\theta }} - {{{y^2}} \over {{{\sin }^2}\theta }} = 1$$
here a = cos$$\theta $$
and b = sin$$\theta $$
We know, eccentricity of the hyperbola is,
$$\sqrt {1 + {{{b^2}} \over {{a^2}}}} $$
$$ \therefore $$ Here eccentricity
(e) = $$\sqrt {1 + {{{{\sin }^2}\theta } \over {{{\cos }^2}\theta }}} $$
Given that,
$$\sqrt {1 + {{{{\sin }^2}\theta } \over {{{\cos }^2}\theta }}} > 2$$
$$ \Rightarrow $$ $$\sqrt {1 + {{\tan }^2}\theta } > 2$$
$$ \Rightarrow $$ 1 + tan2$$\theta $$ > 4
$$ \Rightarrow $$ tan2$$\theta $$ > 3
$$ \Rightarrow $$ tan$$\theta $$ > $$ \pm \sqrt 3 $$
As given $$\theta $$ $$ \in $$ $$\left( {0,{\pi \over 2}} \right)$$
possible value of tan$$\theta $$ > $$\sqrt 3 $$
So, $$\theta $$ can be in the range $${\pi \over 3} < \theta < {\pi \over 2}$$
We know latus ractum (LR) = $${{2{b^2}} \over a}$$
$$ \therefore $$ LR = $${{2{{\sin }^2}\theta } \over {\cos \theta }}$$
= 2 tan$$\theta $$ sin$$\theta $$
We know in the range $${\pi \over 3} < \theta < {\pi \over 2}$$ tan$$\theta $$ and sin$$\theta $$ both are increasing function.
So, at $${\pi \over 3}$$ value of LR will be minimum and at $${\pi \over 2}$$ value of LR will be maximum.
$$ \therefore $$ Minimum value of LR = 2tan$${\pi \over 3}$$ sin$${\pi \over 3}$$
= 2 $$ \times $$ $$\sqrt 3 \times {{\sqrt 3 } \over 2}$$
= 3
Maximum value of LR = 2tan$${\pi \over 2}$$ sin$${\pi \over 2}$$
= 2$$\left( \infty \right) \times 1$$
= $$\infty $$
$$ \therefore $$ Interval of LR = (3, $$\infty $$)
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