Find the locus of intersection of tangents; if the line joining the points of contact to the center be perpendicular in an ellipse.
VerifiedHint: Suppose the standard ellipse with standard equation of chord and now try to form a homogeneous equation which contains both lines passing from origin and joining to point of contact of chord. Then we the condition given in question for these lines (both are perpendicular).
Complete step-by-step solution:
Let equation of ellipse considered \[=\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1-(1)\]
Let tangents are drawn from point P (h, k).
As we know that equation of chord of contact=T=0 w.r.t the points at which tangents are drawn i.e. (h, k) in above diagram.
Equation of chord \[=\dfrac{hx}{{{a}^{2}}}+\dfrac{ky}{{{b}^{2}}}=1-(2)\]
Here, we will use the homogenisation of the curve so that it will pass through (0,0).
Hence, we will form homogeneous equation of curve with the help of chord of contact as follows: -
\[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}={{\left( 1 \right)}^{2}}\because \left( \dfrac{hx}{{{a}^{2}}}+\dfrac{ky}{{{b}^{2}}}=1 \right) \\
\Rightarrow\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}={{\left( \dfrac{hx}{{{a}^{2}}}+\dfrac{ky}{{{b}^{2}}} \right)}^{2}} \\
\Rightarrow\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=\dfrac{{{h}^{2}}{{x}^{2}}}{{{a}^{4}}}+\dfrac{{{k}^{2}}{{y}^{2}}}{{{b}^{4}}}+\dfrac{2hkxy}{{{a}^{2}}{{b}^{2}}} \\
\Rightarrow {{x}^{2}}\left( \dfrac{1}{{{a}^{2}}}-\dfrac{{{h}^{2}}}{{{a}^{4}}} \right)+{{y}^{2}}\left( \dfrac{1}{{{b}^{2}}}-\dfrac{{{k}^{2}}}{{{b}^{2}}} \right)-\dfrac{2hkxy}{{{a}^{2}}{{b}^{2}}}=0-(3) \]
Compare equation (3) with the homogeneous equation of lines \[a{{x}^{2}}+b{{y}^{2}}+2hxy=0\].
And if lines are perpendicular then \[a+b=0.\]
As above the equation is a pair of homogeneous lines passing through origin. i.e., these are lines from chord of contact to centre which are perpendicular.
As we know if lines are perpendicular in homogeneous equation, then coefficient of \[{{x}^{2}}\]+ coefficient of \[{{y}^{2}}\]=0
\[\dfrac{1}{{{a}^{2}}}-\dfrac{{{h}^{2}}}{{{a}^{4}}}+\dfrac{1}{{{b}^{2}}}-\dfrac{{{k}^{2}}}{{{b}^{4}}}=0\]
Replacing h by x and y by k to get locus.
\[\Rightarrow {{a}^{2}}{{b}^{4}}+{{b}^{2}}{{a}^{4}}={{b}^{4}}{{x}^{2}}+{{a}^{4}}{{y}^{2}}\]
\[\Rightarrow{{a}^{4}}{{y}^{2}}+{{b}^{4}}{{x}^{2}}={{a}^{2}}{{b}^{2}}\left( {{a}^{2}}+{{b}^{2}} \right)\]is required locus.
Note: As we know the homogeneous equation of lines passing through origin is \[A{{x}^{2}}+2hxy+B{{y}^{2}}=0\]
Divide the above equation by \[{{x}^{2}}\] both sides, we get
\[B\left( \dfrac{{{y}^{2}}}{{{x}^{2}}} \right)+2h\left( \dfrac{y}{x} \right)+A=0\]
\[\Rightarrow B{{\left( \dfrac{y}{x} \right)}^{2}}+2h\left( \dfrac{y}{x} \right)+A=0\]
As we know that lines passing through origin will have equation as,
\[y={{m}_{1}}x\] and \[y={{m}_{2}}x\]
\[\Rightarrow {{m}_{1}}=\left( \dfrac{y}{x} \right)\] and \[{{m}_{2}}=\left( \dfrac{y}{x} \right)\]
Comparing the values of slopes given above with the quadratic equation in \[\left( \dfrac{y}{x} \right)\], we get that the quadratic equation can be written in form $'m'$ as \[B{{m}^{2}}+2hm+A=0\], which have two roots \[{{m}_{1}}\] and \[{{m}_{2}}\].
\[{{m}_{1}}+{{m}_{2}}=\dfrac{-2h}{B} \\
\Rightarrow {{m}_{1}}{{m}_{2}}=\dfrac{A}{B} \\
\]
If two lines are perpendicular, then\[{{m}_{1}}{{m}_{2}}=-1\]
Hence, \[\dfrac{A}{B}=-1\] or \[A+B=0\] or co-efficient of \[{{x}^{2}}\] + co-efficient of \[{{y}^{2}}\]=0
If two lines are parallel then \[{{m}_{1}}={{m}_{2}}\] or quadratic will have one single root or two equal roots. Therefore, discriminate of quadratic will be zero.
\[{{\left( 2h \right)}^{2}}-4AB=0 \\
\Rightarrow {{h}^{2}}=AB \]
Another approach for this question would be that we can suppose point \[A(a\cos {{\theta }_{1}},b\sin {{\theta }_{1}})\] and point B as \[(a\cos {{\theta }_{2}},b\sin {{\theta }_{2}})\]. We have a relation from given condition i.e. lines joining contact points of tangents to centre are perpendicular. Therefore,
Slope of line AO \[\times \] slope of line BO \[=-1\].
\[\dfrac{b\sin {{\theta }_{1}}}{a\cos {{\theta }_{1}}}\times \dfrac{b\sin {{\theta }_{2}}}{a\cos {{\theta }_{2}}}=-1-(1)\]
Now, write the equations of tangents from point A and B as
\[ \dfrac{x\cos {{\theta }_{1}}}{a}+\dfrac{y\sin {{\theta }_{1}}}{b}=1 \\
\Rightarrow \dfrac{x\cos {{\theta }_{2}}}{a}+\dfrac{y\sin {{\theta }_{2}}}{b}=1 \]
Find intersection of above equations and equate them to h and k, then eliminate \[{{\theta }_{1}}\] and \[{{\theta }_{2}}\] by using equation (1) (mentioned above).
