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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 \[{{x}^{2}}+{{y}^{2}}={{\left( a+b \right)}^{2}}\]

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.

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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