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  • Need explanation for: - If tangents are drawn to the ellipseat all points on the ellipse other than its four vertices th - Co-ordinate geometry - JEE Main

 

If tangents are drawn to the ellipse x^{2}+2y^{2}=2 at all points on the ellipse other than its four vertices then the mid points of the tangents intercepted between the coordinate axes lie on the curve :

  • Option 1)

    \frac{1}{4x^{2}}+\frac{1}{2y^{2}}=1

  • Option 2)

     

    \frac{1}{2x^{2}}+\frac{1}{4y^{2}}=1

     

  • Option 3)

     

    \frac{x^{2}}{4}+\frac{y^{2}}{2}=1

  • Option 4)

     

    \frac{x^{2}}{2}+\frac{y^{2}}{4}=1

Answers (1)

best_answer

 

Standard equation -

\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}= 1
 

- wherein

a\rightarrow Semi major axis

b\rightarrow Semi minor axis

 

 

Locus -

Path followed by a point p(x,y) under given condition (s).

- wherein

It is satisfied by all the points (x,y) on the locus.

 

 

the equation of tangent of P is \frac{x}{\sqrt{2}}\cos \theta +\frac{y\sin \theta }{1}=1

let mid point of AB is (h,k)

Given, (h,k)= \left ( \frac{1}{\sqrt{2}\cos \theta },\frac{1}{2\sin \theta } \right )

as point A\equiv \frac{1}{\sqrt{2}\cos \theta }\Rightarrow \cos \theta =\frac{1}{\sqrt{2}n}

k=\frac{1}{2\sin \theta }\Rightarrow \sin \theta =\frac{1}{2k}

Hence locus of mid point of AB is,

 \cos ^{2}\theta +\sin ^{2}\theta =1=\frac{1}{2h^{2}}+\frac{1}{4k^{2}}=1

putting h=x   k=y

\frac{1}{2x^{2}}+\frac{1}{4y^{2}}=1

 

 


Option 1)

\frac{1}{4x^{2}}+\frac{1}{2y^{2}}=1

Option 2)

 

\frac{1}{2x^{2}}+\frac{1}{4y^{2}}=1

 

Option 3)

 

\frac{x^{2}}{4}+\frac{y^{2}}{2}=1

Option 4)

 

\frac{x^{2}}{2}+\frac{y^{2}}{4}=1

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