Question

Find the locus of intersection of tangents which meet at a given angle \[\alpha \]with ellipse \[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\].

Answer 1

Hint: Use the standard equation of tangent for ellipse for tangent \[y=mx\pm \sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}\], then form a quadratic in ‘m’ and use formula \[\tan \theta =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|\]to get locus.

Complete step-by-step answer:
We have considered ellipse equation as \[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\left\{ \because a>b \right\}\]

Let the point of intersection be P (h, k).
As we know,
Equation to the tangent \[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\]in slope form is
\[y=mx\pm \sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}\]
As, this equation will pass through (h, k).
Hence,
\[k=mh\pm \sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}\]
\[k-mh=\pm \sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}\]
Squaring both sides
\[\begin{align}
  & {{\left( k-mh \right)}^{2}}={{a}^{2}}{{m}^{2}}+{{b}^{2}} \\
 & {{k}^{2}}+{{m}^{2}}{{h}^{2}}-2kmh={{a}^{2}}{{m}^{2}}+{{b}^{2}} \\
 & \left( {{h}^{2}}-{{a}^{2}} \right){{m}^{2}}-2mhk+{{k}^{2}}-{{b}^{2}}=0-(1)\left\{ \because {{\left( APB \right)}^{2}}={{A}^{2}}+{{B}^{2}}+2AB \right\} \\
\end{align}\]
This equation is quadratic in m and has two roots. Let’s suppose \[{{m}_{1}}\]and \[{{m}_{2}}\]which are slopes of \[{{T}_{1}}\]and \[{{T}_{2}}\]tangents shown in diagram: -
As, if we have quadratic \[A{{X}^{2}}+BX+C=0\]
Then sum of roots \[=\dfrac{-B}{A}\]
Product of roots \[=\dfrac{C}{A}\]
Hence, from the equation (1)
\[\begin{align}
  & {{m}_{1}}+{{m}_{2}}=\dfrac{2hk}{{{h}^{2}}-{{a}^{2}}} \\
 & {{m}_{1}}.{{m}_{2}}=\dfrac{{{k}^{2}}-{{b}^{2}}}{{{h}^{2}}-{{a}^{2}}} \\
\end{align}\]
As, \[\tan \alpha =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|\]

If slope of two lines are given then \[\tan \alpha =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|\]as \[\alpha \]is angle between two lines.
\[\begin{align}
  & {{\tan }^{2}}\alpha =\dfrac{{{\left( {{m}_{1}}-{{m}_{2}} \right)}^{2}}}{{{\left( 1+{{m}_{1}}{{m}_{2}} \right)}^{2}}} \\
 & {{\tan }^{2}}\alpha =\dfrac{{{\left( {{m}_{1}}+{{m}_{2}} \right)}^{2}}-4{{m}_{1}}{{m}_{2}}}{{{\left( 1+{{m}_{1}}{{m}_{2}} \right)}^{2}}}\left\{ \because {{\left( a-b \right)}^{2}}={{\left( a+b \right)}^{2}}-4ab \right\} \\
 & {{\tan }^{2}}\alpha =\dfrac{{{\left( \dfrac{2hk}{{{h}^{2}}-{{a}^{2}}} \right)}^{2}}-4\left( \dfrac{{{k}^{2}}-{{b}^{2}}}{{{h}^{2}}-{{a}^{2}}} \right)}{{{\left( 1+\dfrac{{{k}^{2}}-{{b}^{2}}}{{{h}^{2}}-{{a}^{2}}} \right)}^{2}}} \\
 & {{\tan }^{2}}\alpha =\dfrac{4{{h}^{2}}{{k}^{2}}-4\left( {{h}^{2}}-{{a}^{2}} \right)\left( {{k}^{2}}-{{b}^{2}} \right)}{{{\left( {{h}^{2}}+{{k}^{2}}-{{a}^{2}}-{{b}^{2}} \right)}^{2}}} \\
\end{align}\]
Replacing (h, k) by (x, y) to get locus: -
\[{{\left( {{x}^{2}}+{{y}^{2}}-{{a}^{2}}-{{b}^{2}} \right)}^{2}}=4{{\cot }^{2}}\alpha \left( {{x}^{2}}{{b}^{2}}+{{a}^{2}}{{y}^{2}}-{{a}^{2}} \right)\]is required locus.
Note: Using formula \[\tan \alpha =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|\]and writing relations from tangent equation \[y=mx\pm \sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}\]or \[{{\left( y-mx \right)}^{2}}={{a}^{2}}{{m}^{2}}+{{b}^{2}}\]which is quadratic in m and using relations of roots with coefficients of quadratic is a key point of this equation.
We can use the direct formula of tangent of ellipse i.e. \[y=mx\pm \sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}\]standard equation \[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\]which can be proved by following approach: -

Now, y=mx + c and \[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\]have only one intersection point (touching the ellipse). So, if we substitute y=mx + c in \[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\]at place of y then we will get quadratic in x which should have one solution (as tangent and ellipse have only one intersection point). Then we will get \[c=\pm \sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}\]by making decrement of that quadratic to 0.
Second approach to the solution can be: -
We can suppose A and B points as a parametric coordinates of ellipse as \[A\left( a\sin {{\theta }_{1}},b\cos {{\theta }_{1}} \right)\And B\left( a\sin {{\theta }_{2}},b\cos {{\theta }_{2}} \right)\]and write tangent equations from A and B as ‘T=0’ or
\[\begin{align}
  & \dfrac{a\sin {{\theta }_{1}}}{a}+\dfrac{y\cos {{\theta }_{1}}}{b}=1 \\
 & \dfrac{a\sin {{\theta }_{2}}}{a}+\dfrac{y\cos {{\theta }_{2}}}{b}=1 \\
\end{align}\]
And then find the intersection of above two tangent and trying to eliminate \[{{\theta }_{1}}\And {{\theta }_{2}}\]by using the given condition with using same formula \[\tan \alpha =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|\].
Where \[{{m}_{1}}=\dfrac{\dfrac{-\sin {{\theta }_{1}}}{a}}{\dfrac{\cos {{\theta }_{1}}}{b}}=\dfrac{-b\sin {{\theta }_{1}}}{a\cos {{\theta }_{1}}}\]
\[{{m}_{2}}=\dfrac{\dfrac{-\sin {{\theta }_{2}}}{a}}{\dfrac{\cos {{\theta }_{2}}}{b}}=\dfrac{-b\sin {{\theta }_{2}}}{a\cos {{\theta }_{2}}}\]