Recall Properties Of Ellipse

Note that: If an tangent is drawn to the ellipse then it has following Properties:

$1)-\quad \quad { P }_{ 1 }\quad ,\quad d\quad ,\quad { P }_{ 2 }\quad \longrightarrow AP\\ 2)-\quad \quad { V }_{ 1 }\quad ,\quad d\quad ,\quad { V }_{ 2 }\quad \longrightarrow AP\\ 3)-\quad \quad { P }_{ 1 }{ P }_{ 2 }={ b }^{ 2 }$ .

Where ${ V }_{ 1 }\quad \& \quad { V }_{ 2 }\quad$ . are perpendiculars (SHORTEST DISTANCE) from vertex of major axis of ellipse to the same tangent. However This is not used in this question but i posted it for only Information Purpose.

(Yes You can prove it but i skip this.)

So $\left| { P }_{ 1 }\quad -\quad { P }_{ 2 } \right| =\sqrt { { ({ P }_{ 1 }\quad +\quad { P }_{ 2 }) }^{ 2 }-4{ P }_{ 1 }{ P }_{ 2 } } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \sqrt { { (2d) }^{ 2 }-4{ P }_{ 1 }{ P }_{ 2 } } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\sqrt { { 4d }^{ 2 }-4{ P }_{ 1 }{ P }_{ 2 } } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =2\sqrt { { d }^{ 2 }-{ b }^{ 2 } }$ .

NOTE: Here i used the Fact that if :

${ P }_{ 1 }\quad ,\quad d\quad ,\quad { P }_{ 2 }\quad \longrightarrow AP$ .

then

${ P }_{ 1 }\quad +\quad { P }_{ 2 }\quad =\quad 2d$ .

Now Put the value and get The Answer.

$If$ $Anyone$ $Knows$ $More$ $Properties$ $Related$ $To$ $Tangent$ $of$ $Ellipse$ $Then$ $Please$ $Share$ $it$ $with$ $us$ .

We know that foot of the perpendicular from the focus to the tangents of the conic lies on the auxiliary circle.

Now the auxiliary circle of the ellipse has a radius of $a$ and is centered at the origin.

So the information given to us points out that the radius of the auxiliary circle is $a=5$ .

So we can find out the eccentricity of the ellipse which comes out to be $0.8$ .

Again using a bit of common sense( or just by using parametric equations of $acos(t)$ and $bsin(t)$ and using distance formula to find distance of the tangent from the origin and forming the equation to get equate that t = 0 or pi) we can easily see that the tangent must have been drawn at the vertex and it is parallel to y axis.

So P1 - P2 . just comes out to be the difference between the distance of the line $x=5$ from the foci or $x=-5$ from both foci which comes out to be 8 which is the answer