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Question: If two tangents to an ellipse and one of its foci be given, prove that the locus of its centre is a ...

If two tangents to an ellipse and one of its foci be given, prove that the locus of its centre is a straight line.

Explanation

Solution

Hint: Consider T as the tangent drawn to the ellipse. Use the equation to find the polar form of tangent. Take focus ‘S’ of the ellipse is (ae, 0) and for ss' (ae,0)\left( -ae,0 \right). Find the polar form of Tangents to prove the locus of its centre is a straight line.

Complete step by step answer:
Let us consider TT as the tangent that is drawn to the ellipse. TPTP and TQTQ are the two tangents from TT, which has their focus at S.S. The focus SS of the ellipse is (ae,0)\left( ae,0 \right) and for S1S_1 (ae,0)\left( -ae,0 \right).
Hence we need to prove that QTS1=PTS\angle QT{{S}_{1}}=\angle PTS ( from figure).

The centre CC of the ellipse is midway between S and S1{{S}_{1}} , which are the focus.
Let the coordinate of any point be (h,k)\left( h,k \right) and equation to ellipse is
x2a2+y2b2=1............(1)\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1............\left( 1 \right)
To find the polar form of tangent we have the equation xx1a2+yy1b2=1..............(2)\dfrac{x{{x}_{1}}}{{{a}^{2}}}+\dfrac{y{{y}_{1}}}{{{b}^{2}}}=1..............\left( 2 \right)
(h,k)=(x1,y1)\left( h,k \right)=\left( {{x}_{1}},{{y}_{1}} \right)
\therefore Substituting values in (2)xha2+ykb2=1...........(3)\left( 2 \right)\Rightarrow \dfrac{xh}{{{a}^{2}}}+\dfrac{yk}{{{b}^{2}}}=1...........\left( 3 \right)
By hypothesis 1h2a2+k2b2=c\Rightarrow \dfrac{1}{\sqrt{\dfrac{{{h}^{2}}}{{{a}^{2}}}+\dfrac{{{k}^{2}}}{{{b}^{2}}}}}=c is the length of perpendicular from centre (0,0)\left( 0,0 \right) is given this
On squaring and cross multiplying we get
12=h2a4+k2b4\dfrac{1}{2}=\dfrac{{{h}^{2}}}{{{a}^{4}}}+\dfrac{{{k}^{2}}}{{{b}^{4}}}
Now generalizing the locus of (h,k)\left( h,k \right) , we get x2a4+y2b4=1c2\dfrac{{{x}^{2}}}{{{a}^{4}}}+\dfrac{{{y}^{2}}}{{{b}^{4}}}=\dfrac{1}{{{c}^{2}}}

We got the locus of its centre, so when S1{{S}_{1}} moves on TS1,CT{{S}_{1}}, C will move parallel to TS1T{{S}_{1}} so that they will bisect the line SS1S{{S}_{1}}. Therefore we were able to prove that the locus of centre CC is a straight line.

Note: Remember that two tangents drawn from any point to an ellipse subtend equal angle at the focus.