Question

Jee level question
Chapter:- Coordinate geometry ​

Answer 1

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

Given that,

$\rm :\longmapsto\: \alpha, \beta \: lies \: on \: the \: parabola \: {y}^{2} = x$

$\rm :\implies\: { \beta }^{2} = \alpha - - - - (1)$

Now, we know that Equation of tangent to parabola y² = 4ax at a point ( p, q ) is given by yq = 2a( x + p ).

So,

$\rm :\longmapsto\:Equation \: of \: tangent \: to \: {y}^{2} = x \: at \: ( \alpha , \beta ) \: is \:$

$\rm :\longmapsto\:y \beta = \dfrac{x + \alpha }{2}$

$\rm :\longmapsto\:y \beta = \dfrac{x + { \beta }^{2} }{2}$

$\red{\bigg \{ \because \: \alpha = { \beta }^{2} \bigg \}}$

Now, above equation can be rewritten as

$\rm :\longmapsto\:y = \dfrac{x + { \beta }^{2} }{2 \beta }$

$\bf\implies \:y = \dfrac{1}{2 \beta }x + \dfrac{ \beta }{2} - - - (2)$

Now,

Given that

$\rm :\longmapsto\:y = \dfrac{1}{2 \beta }x + \dfrac{ \beta }{2} \: is \: tangent \: to \: {x}^{2} + {2y}^{2} = 1$

can be rewritten as

$\rm :\longmapsto\:y = \dfrac{1}{2 \beta }x + \dfrac{ \beta }{2} \: is \: tangent \: to \: {x}^{2} + \dfrac{ {y}^{2} }{ \: \: \dfrac{1}{2} \: \: \: } = 1$

We know,

$\red{\rm :\longmapsto\:y = mx + x \: is \: tangent \: to \: ellipse \: \dfrac{ {x}^{2} }{ {a}^{2} } + \dfrac{ {y}^{2} }{ {b}^{2} } = 1 \: then} \\ \blue{\rm :\longmapsto\:c \: = \: \pm \: \sqrt{ {a}^{2} \: {m}^{2} \: + \: {b}^{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }$

So, here

$\rm :\longmapsto\:m = \dfrac{1}{2 \beta }$

$\rm :\longmapsto\:c = \dfrac{ \beta }{2}$

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

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

So, on substituting the values, we get

$\rm :\longmapsto\:\dfrac{ \beta }{2} = \pm \: \sqrt{1 \times \dfrac{ { 4 }}{ { \beta }^{2} } + \dfrac{1}{2} }$

$\rm :\longmapsto\:\dfrac{ \beta }{2} = \pm \: \sqrt{ \dfrac{ { 4 }}{ { \beta }^{2} } + \dfrac{1}{2} }$

On squaring both sides, we get

$\rm :\longmapsto\:\dfrac{ { \beta }^{2} }{4} = \dfrac{ { 4 }}{ { \beta }^{2} } + \dfrac{1}{2}$

$\rm :\longmapsto\:\dfrac{ { \beta }^{2} }{4} - \dfrac{ { 4 }}{ { \beta }^{2} } - \dfrac{1}{2} = 0$

$\rm :\longmapsto\: { \beta }^{4} - {2 \beta }^{2} - 1 = 0$

$\rm :\longmapsto\: { \beta }^{4} - {2 \beta }^{2} = 1$

$\rm :\longmapsto\: { \beta }^{4} - {2 \beta }^{2} + 1 = 1 + 1$

$\rm :\longmapsto\: {( { \beta }^{2} - 1)}^{2} = 2$

$\rm :\longmapsto\: {{ \beta }^{2} - 1}= \sqrt{2} \: \: \: \: as \: \beta > 0$

$\rm :\implies\: { \beta }^{2} = \sqrt{2} + 1$

$\bf :\implies\: \alpha = \sqrt{2} + 1 \: \: \: \: \: \: \: as \: { \beta }^{2} = \alpha$

Hence,

$\: \: \: \: \: \: \: \: \: \: \: \boxed{ \bf{ \: Option \: (a) \: is \: correct \: \: }}$