Question

Practice set 1( mathamatics)
Important for jee mains exam ​

Answer 1

$\huge \frac{1}{ \sqrt{2} }$

Answer 2

$\large\underline{\sf{Solution-}}$

Let we assume that for the ellipse

$\rm :\longmapsto\:\dfrac{ {x}^{2} }{ {a}^{2} } + \dfrac{ {y}^{2} }{ {b}^{2} } = 1 \: \: with \: a > b$

So, end point of minor axis be (0, b) and (0, - b).

Let assume that the one end point of minor axis be represented as A(0, b) and other as F (0, - b).

Now,

We know that,

Coordinates of Focus is given by (ae, 0) and (- ae, 0).

where, e is eccentricity of ellipse and is given by

$\blue{\boxed{ \sf{ \: {b}^{2} = {a}^{2}(1 - {e}^{2})}}}$

$\rm :\implies\: {a}^{2} = {b}^{2} + {a}^{2} {e}^{2} - - - (1)$

Now,

Let assume that coordinates of focus be represented by B and C.

That means,

Coordinates of B be ( - ae, 0 )

and

Coordinates of C be ( ae, 0 )

Now,

We further know that,

Distance between focus = 2ae

So, it implies, BC = 2ae.

Now,

$\rm :\longmapsto\:AB = \sqrt{ {(0 + ae)}^{2} + (b - 0)^{2} }$

$\rm :\longmapsto\:AB = \sqrt{ { a^{2} e}^{2} + b^{2} }$

$\rm :\longmapsto\:AB = \sqrt{{a^{2 }}} \: \: \: \: \{ \: using \: (1) \}$

$\bf\implies \:AB = a$

Again,

$\rm :\longmapsto\:AC= \sqrt{ {(0 - ae)}^{2} + (b - 0)^{2} }$

$\rm :\longmapsto\:AC= \sqrt{ {(ae)}^{2} + b^{2} }$

$\rm :\longmapsto\:AC= \sqrt{ {a^{2} e}^{2} + b^{2} }$

$\rm :\longmapsto\:AC = \sqrt{{a^{2 }}} \: \: \: \: \{ \: using \: (1) \}$

$\bf\implies \:AC = a$

So, Now In right angle triangle ABC, we have

$\rm :\longmapsto\:AB = a$

$\rm :\longmapsto\:AC = a$

$\rm :\longmapsto\:BC = 2ae$

$\rm :\longmapsto\: \angle \: BAC \: = \: 90 \degree$

Using Pythagoras Theorem,

$\rm :\longmapsto\: {AB}^{2} + {AC}^{2} = {BC}^{2}$

$\rm :\longmapsto\: {a}^{2} + {a}^{2} = 4{a}^{2} {e}^{2}$

$\rm :\longmapsto\: 2{a}^{2} = 4{a}^{2} {e}^{2}$

$\rm :\longmapsto\: 1= 2{e}^{2}$

$\rm :\longmapsto\: {e}^{2} = \dfrac{1}{2}$

$\bf\implies \:e = \dfrac{1}{ \sqrt{2} }$

Hence,

• Option (c) is correct.