Question

From point P (8, 27) tangents PQ and PR are drawn to the ellipse,$\dfrac{{{{\text{x}}^2}}}{4} + \dfrac{{{{\text{y}}^2}}}{9} = 1$, then the angle subtended by QR at the origin is
${\text{A}}{\text{. ta}}{{\text{n}}^{ - 1}}\dfrac{{2\sqrt 6 }}{{65}} \\\ {\text{B}}{\text{. ta}}{{\text{n}}^{ - 1}}\dfrac{{4\sqrt 6 }}{{65}} \\\ {\text{C}}{\text{. ta}}{{\text{n}}^{ - 1}}\dfrac{{8\sqrt 2 }}{{65}} \\\ {\text{D}}{\text{. None of these}} \\\$

Answer 1

In order to find the angle made by QR at the origin, we first find out the equation of chord of contact QR to the ellipse using the point P and the ellipse equation. Then we find the equation of the pair of lines passing through the points Q and R. Now the angle subtended by the line QR is nothing but the angle between the lines, we find it using its formula.

Complete step by step answer:
Given Data,
Point P (8, 27)
Ellipse equation $\dfrac{{{{\text{x}}^2}}}{4} + \dfrac{{{{\text{y}}^2}}}{9} = 1$
Tangents from point P, PQ and PR are drawn to the ellipse

We know the equation of a chord of contact made by the tangents coming from a point of the form (a, b) to the ellipse of the form $\dfrac{{{{\text{x}}^2}}}{{\text{p}}} + \dfrac{{{{\text{y}}^2}}}{{\text{q}}} = {\text{c}}$is given by,
Chord:$\dfrac{{{\text{ax}}}}{{\text{p}}} + \dfrac{{{\text{by}}}}{{\text{q}}} = {\text{c}}$
Therefore the equation of the chord of contact QR to the ellipse$\dfrac{{{{\text{x}}^2}}}{4} + \dfrac{{{{\text{y}}^2}}}{9} = 1$from point P is given by,
Chord equation QR - $\dfrac{{{\text{8x}}}}{4} + \dfrac{{{\text{27y}}}}{9} = 1 \Rightarrow {\text{2x + 3y = 1 - - - - - }}\left( 1 \right)$

Now the equation of pair of lines passing through the origin and the point Q and R can be determined using the equation of ellipse and the equation of the chord which joins these points, i.e. QR as follows:
Equation of pair of lines,
$\Rightarrow \left( {\dfrac{{{{\text{x}}^2}}}{4} + \dfrac{{{{\text{y}}^2}}}{9}} \right) = {\left( {{\text{2x + 3y}}} \right)^2} \\\ \Rightarrow {\text{9}}{{\text{x}}^2} + 4{{\text{y}}^2} = 36\left( {{\text{4}}{{\text{x}}^2} + 12{\text{xy + 9}}{{\text{y}}^2}} \right) \\\ \Rightarrow 135{{\text{x}}^2} + 432{\text{xy + 320}}{{\text{y}}^2} = 0 \\\$

Now we know angle between a pair of lines of the form, ${\text{a}}{{\text{x}}^2} + {\text{k}}{\text{.xy + b}}{{\text{y}}^2} = 0$is given by
$\theta {\text{ = ta}}{{\text{n}}^{ - 1}}\left( {\dfrac{{2\sqrt {{{\left( {\dfrac{{\text{k}}}{2}} \right)}^2} - {\text{a}}{\text{.b}}} }}{{{\text{a + b}}}}} \right)$

The angle subtended by the chord QR at the origin is nothing but the angle made by the lines passing through the origin and point Q and R respectively.
Therefore the angle between the pair of lines, $135{{\text{x}}^2} + 432{\text{xy + 320}}{{\text{y}}^2} = 0$ or subtended by the chord QR is given by
$\Rightarrow \theta {\text{ = ta}}{{\text{n}}^{ - 1}}\left( {\dfrac{{2\sqrt {{{\left( {\dfrac{{432}}{2}} \right)}^2} - 135.{\text{320}}} }}{{135 + 320}}} \right) \\\ \Rightarrow \theta {\text{ = ta}}{{\text{n}}^{ - 1}}\left( {\dfrac{{2\sqrt {{{\left( {216} \right)}^2} - 2700} }}{{455}}} \right) \\\ \Rightarrow \theta {\text{ = ta}}{{\text{n}}^{ - 1}}\left( {\dfrac{{8\sqrt {216} }}{{455}}} \right) \\\ \Rightarrow \theta {\text{ = ta}}{{\text{n}}^{ - 1}}\dfrac{{48\sqrt 6 }}{{455}} \\\$

Hence the angle subtended by QR at the origin is${\text{ta}}{{\text{n}}^{ - 1}}\dfrac{{48\sqrt 6 }}{{455}}$.
Option D is the correct answer.

Note: In order to solve this type of problems the key is to know the concepts of geometry like ellipse, pair of lines, equation of tangents from a point and equation of chords. We use the formulae of all these respective concepts to find the equation of the pair of lines and find the angle between them.
Understanding that the angle made by the chord with the origin is nothing but the angle between the lines because they pass through the origin and the individual points is the most important step of this problem.
We have to watch out while performing the calculation part as it can get tricky and people tend to make errors here.