If two tangents to an ellipse and one of its foci be given, prove that the locus of its centre is a straight line.
Answer
627.9k+ views
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 $s'$ $\left( -ae,0 \right)$. 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 $T$ as the tangent that is drawn to the ellipse. $TP$ and $TQ$ are the two tangents from $T$, which has their focus at $S.$ The focus $S$ of the ellipse is $\left( ae,0 \right)$ and for $S_1$ $\left( -ae,0 \right)$.
Hence we need to prove that $\angle QT{{S}_{1}}=\angle PTS$ ( from figure).
The centre $C$ of the ellipse is midway between S and ${{S}_{1}}$ , which are the focus.
Let the coordinate of any point be $\left( h,k \right)$ and equation to ellipse is
$\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1............\left( 1 \right)$
To find the polar form of tangent we have the equation $\dfrac{x{{x}_{1}}}{{{a}^{2}}}+\dfrac{y{{y}_{1}}}{{{b}^{2}}}=1..............\left( 2 \right)$
$\left( h,k \right)=\left( {{x}_{1}},{{y}_{1}} \right)$
$\therefore $ Substituting values in $\left( 2 \right)\Rightarrow \dfrac{xh}{{{a}^{2}}}+\dfrac{yk}{{{b}^{2}}}=1...........\left( 3 \right)$
By hypothesis $\Rightarrow \dfrac{1}{\sqrt{\dfrac{{{h}^{2}}}{{{a}^{2}}}+\dfrac{{{k}^{2}}}{{{b}^{2}}}}}=c$ is the length of perpendicular from centre $\left( 0,0 \right)$ is given this
On squaring and cross multiplying we get
$\dfrac{1}{2}=\dfrac{{{h}^{2}}}{{{a}^{4}}}+\dfrac{{{k}^{2}}}{{{b}^{4}}}$
Now generalizing the locus of $\left( h,k \right)$ , we get $\dfrac{{{x}^{2}}}{{{a}^{4}}}+\dfrac{{{y}^{2}}}{{{b}^{4}}}=\dfrac{1}{{{c}^{2}}}$
We got the locus of its centre, so when ${{S}_{1}}$ moves on $T{{S}_{1}}, C$ will move parallel to $T{{S}_{1}}$ so that they will bisect the line $S{{S}_{1}}$. Therefore we were able to prove that the locus of centre $C$ is a straight line.
Note: Remember that two tangents drawn from any point to an ellipse subtend equal angle at the focus.
Complete step by step answer:
Let us consider $T$ as the tangent that is drawn to the ellipse. $TP$ and $TQ$ are the two tangents from $T$, which has their focus at $S.$ The focus $S$ of the ellipse is $\left( ae,0 \right)$ and for $S_1$ $\left( -ae,0 \right)$.
Hence we need to prove that $\angle QT{{S}_{1}}=\angle PTS$ ( from figure).
The centre $C$ of the ellipse is midway between S and ${{S}_{1}}$ , which are the focus.
Let the coordinate of any point be $\left( h,k \right)$ and equation to ellipse is
$\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1............\left( 1 \right)$
To find the polar form of tangent we have the equation $\dfrac{x{{x}_{1}}}{{{a}^{2}}}+\dfrac{y{{y}_{1}}}{{{b}^{2}}}=1..............\left( 2 \right)$
$\left( h,k \right)=\left( {{x}_{1}},{{y}_{1}} \right)$
$\therefore $ Substituting values in $\left( 2 \right)\Rightarrow \dfrac{xh}{{{a}^{2}}}+\dfrac{yk}{{{b}^{2}}}=1...........\left( 3 \right)$
By hypothesis $\Rightarrow \dfrac{1}{\sqrt{\dfrac{{{h}^{2}}}{{{a}^{2}}}+\dfrac{{{k}^{2}}}{{{b}^{2}}}}}=c$ is the length of perpendicular from centre $\left( 0,0 \right)$ is given this
On squaring and cross multiplying we get
$\dfrac{1}{2}=\dfrac{{{h}^{2}}}{{{a}^{4}}}+\dfrac{{{k}^{2}}}{{{b}^{4}}}$
Now generalizing the locus of $\left( h,k \right)$ , we get $\dfrac{{{x}^{2}}}{{{a}^{4}}}+\dfrac{{{y}^{2}}}{{{b}^{4}}}=\dfrac{1}{{{c}^{2}}}$
We got the locus of its centre, so when ${{S}_{1}}$ moves on $T{{S}_{1}}, C$ will move parallel to $T{{S}_{1}}$ so that they will bisect the line $S{{S}_{1}}$. Therefore we were able to prove that the locus of centre $C$ is a straight line.
Note: Remember that two tangents drawn from any point to an ellipse subtend equal angle at the focus.
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

Which among the following are examples of coming together class 11 social science CBSE

