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 \[{{x}^{2}}+{{y}^{2}}={{\left( a+b \right)}^{2}}\].
Answer
363.9k+ views
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.
Last updated date: 27th Sep 2023
•
Total views: 363.9k
•
Views today: 6.63k
Recently Updated Pages
What do you mean by public facilities

Difference between hardware and software

Disadvantages of Advertising

10 Advantages and Disadvantages of Plastic

What do you mean by Endemic Species

What is the Botanical Name of Dog , Cat , Turmeric , Mushroom , Palm

Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Why are resources distributed unequally over the e class 7 social science CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Briefly mention the contribution of TH Morgan in g class 12 biology CBSE

What is the past tense of read class 10 english CBSE
