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Any ordinate \[NP\] of an ellipse meets the auxiliary circle in \[Q\]; prove that the locus of the intersection of the normals at \[P\] and \[Q\] is the circle \[{{x}^{2}}+{{y}^{2}}={{\left( a+b \right)}^{2}}\].

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Hint: Write the equation of the normals to ellipse and circle and satisfy \[\left( h,k \right)\] in them.

 We are given an ellipse with ordinate \[NP\] which meets the auxiliary circle at \[Q\].
We have to prove the locus of intersection of normals at \[P\] and \[Q\] is the circle \[{{x}^{2}}+{{y}^{2}}={{\left( a+b \right)}^{2}}\]


Let \[\left( h,k \right)\] be the point at which normals at \[P\] and \[Q\] meet. Here, \[NP\] is ordinary which intersects the auxiliary circle at \[Q\].
Let the ellipse be \[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\]
Since, \['a'\] is the radius of the auxiliary circle with a center \[\left( 0,0 \right)\].
Therefore, we get the equation of the auxiliary circle as –
\[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\]
We know that any general point \[P\] on the ellipse is given by \[\left( x,y \right)=\left( a\cos \theta ,b\sin \theta \right)\]
Also, we know that normal of the ellipse at a point \[P\left( a\cos \theta ,b\sin \theta \right)\] is given by
\[\dfrac{ax}{\cos \theta }-\dfrac{by}{\sin \theta }={{a}^{2}}-{{b}^{2}}\]
Since, this normal passes through \[\left( h,k \right)\], then
We get, \[\dfrac{ah}{\cos \theta }-\dfrac{bk}{\sin \theta }={{a}^{2}}-{{b}^{2}}.....\left( i \right)\]
We know the normal to the circle always passes through its center.
Therefore, normal at a point \[Q\] will pass through the center of the auxiliary circle \[\left( 0,0 \right)\]. So the normal at \[Q\] is \[OQ\] which makes an angle \[\theta \] with the \[x\] axis.
We know that equation of any line passing through \[\left( 0,0 \right)\] and the slope \['m'\] is given by
\[\Rightarrow y=mx\]
Therefore, the equation of \[OQ\] is also
\[\Rightarrow y=mx\]
Since \[NP\] is ordinate which meets the auxiliary circle at \[Q\]. Therefore, \[OQ\] will make an angle \[\theta \] with the \[x\] axis.
So, \[m=\tan \theta \]
Hence, we get the equation of \[OQ\] as
\[\Rightarrow y=\left( \tan \theta \right)x\]
\[OQ\] also passes through \[P\left( h,k \right)\], then
\[\Rightarrow k=\left( \tan \theta \right)h\]
Or, \[\tan \theta =\dfrac{k}{h}\]


Therefore, we get \[\sin \theta =\dfrac{k}{\sqrt{{{h}^{2}}+{{k}^{2}}}}\] and \[\cos \theta =\dfrac{h}{\sqrt{{{h}^{2}}+{{k}^{2}}}}\]
Putting the values of \[\sin \theta \] and \[\cos \theta \] in equation \[\left( i \right)\], we get
\[\dfrac{ah\left( \sqrt{{{h}^{2}}+{{k}^{2}}} \right)}{h}-\dfrac{bk\left( \sqrt{{{h}^{2}}+{{k}^{2}}} \right)}{k}={{a}^{2}}-{{b}^{2}}\]
\[\Rightarrow a\left( \sqrt{{{h}^{2}}+{{k}^{2}}} \right)-b\left( \sqrt{{{h}^{2}}+{{k}^{2}}} \right)={{a}^{2}}-{{b}^{2}}\]
By taking \[\sqrt{{{h}^{2}}+{{k}^{2}}}\] common, we get
\[\Rightarrow \left( \sqrt{{{h}^{2}}+{{k}^{2}}} \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}\]
\[\Rightarrow \left( \sqrt{{{h}^{2}}+{{k}^{2}}} \right)=\dfrac{\left( {{a}^{2}}-{{b}^{2}} \right)}{\left( a-b \right)}\]
Since we know that \[{{x}^{2}}-{{y}^{2}}=\left( x-y \right)\left( x+y \right)\]
We get, \[\Rightarrow \left( \sqrt{{{h}^{2}}+{{k}^{2}}} \right)=\dfrac{\left( a-b \right)\left( a+b \right)}{\left( a-b \right)}\]
We get, \[\sqrt{{{h}^{2}}+{{k}^{2}}}=\left( a+b \right)\]
By squaring both sides, we get
\[\Rightarrow {{h}^{2}}+{{k}^{2}}={{\left( a+b \right)}^{2}}\]
To get the locus, replace \[h\] and \[k\] by \[x\] and \[y\] respectively.
\[\Rightarrow {{x}^{2}}+{{y}^{2}}={{\left( a+b \right)}^{2}}\]
Hence Proved

Note: In questions involving ellipse, always take the general points in terms of \[\theta \] to easily solve the questions. Note the normal to circle always passes through its center.
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