Question

Let 'd' be the perpendicular distance from the centre of the ellipse

$\frac{ {x}^{2} }{ {a}^{2} } + \frac{ {y}^{2} }{ {b}^{2} } = 1$

to the tangent drawn at a point P on the ellipse.

If $F_1\:andF_2\:$ are the two focii of the ellipse, then show that,

${(PF_1-PF_2)}^{2}=4{a}^{2}(1-\frac{{b}^{2}}{{d}^{2}})$


✔️✔️Proper solution needed✔️✔️​

Answer 1

ANSWER

Plz refer to the attachment for diagram.

Step By Step Explanation

Let the coordinates of P be ( $aCos\theta , bSin\theta$ )

Now, equation of tangent at P is >

$\dfrac{x}{a}Cos\theta + \dfrac{y}{b}Sin \theta$ = 1

Now,

d = $\dfrac{0+0-1}{ \sqrt(\dfrac{cos^{2}\theta}{a^{2} } + \dfrac{Sin^{2} \theta}{b^{2} }) }$

=>$\dfrac{1}{d^{2} }$

= $\dfrac{1}{\dfrac{Cos^{2} }{a^{2} } + \dfrac{Sin^{2} }{b^{2} } }$

Now,

RHS

$4a^{2}(1 - \dfrac {b^{2} }{d^{2} } )$

=>$4a^{2} - \dfrac{4a^{2}b^{2} } {d^{2} }$

=>$( 4a^{2} - 4a^{2}b^{2} ) ( \dfrac{Cos^{2}\theta}{a^{2} } + \dfrac{Sin^{2}\theta}{b^{2} } )$

=>$4a^{2} - 4b^{2}Cos^{2}\theta - 4a^{2}Sin^{2}\theta$

=> $4a^{2}(1 - Sin^{2}\theta) -4b^{2}Cos^{2}\theta$

=> $4a^{2}Cos^{2}\theta - 4 b^{2} Cos^{2}\theta$

=> $4Cos^{2}\theta(a^{2} - b^{2})$

=>$4Cos^{2}\theta\times a^{2}e^{2}$

Again,

$PF_{1} = e|aCos\theta + \dfrac{a}{e}|$

=> $PF_{1} = a|eCos\theta + 1|$

=> $PF_{1} = a(eCos\theta + 1)$

Similarly we have >

$PF_{2} =a(1 - eCos\theta)$

Now,

RHS

$(PF_{1} - PF_{2} )^{2}$

=> $( a(eCos\theta + 1) - a(1 - eCos\theta) )^{2}$

=> $(2ae Cos\theta)^{2}$

=>$4a^{2}e^{2}Cos^{2}$

Hence , LHS = RHS

Proved