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Locus of midpoints of chords of the ellipse \frac{x^2}{2}+y^2=1 which are tangents to the ellipse x^2+\frac{y^2}{2}=1 is 

Option: 1

(2y^2+x^2)^2={x^2+8y^2}


Option: 2

(2x^2+y^2)^2={x^2+8y^2}


Option: 3

(2y^2-x^2)^2={x^2+8y^2}


Option: 4

(2x^2-y^2)^2={x^2+8y^2}


Answers (1)

best_answer

 

 

Diameter of Ellipse -

Diameter of Ellipse:

The locus of the mid-points of a system of parallel chords of an ellipse is called a diameter and the point where the diameter intersects the ellipse is called the vertex of the diameter.

\\\text{Let (h, k) be the mid-point of the chord y=m x+c of the ellipse}\\\mathrm{\text { } \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\;,\;then}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;T=S_1\;\;\;\;[\text{equation\;of\;chord\;bisected at given point}]}\\\\\mathrm{\Rightarrow \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{x h}{a^{2}}+\frac{y k}{b^{2}}=\frac{h^{2}}{a^{2}}+\frac{k^{2}}{b^{2}}}\\\\\mathrm{\Rightarrow \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;k=-\frac{b^{2} h}{a^{2} m}}\\\\\text{Hence, the locus of the mid-point is }y=-\frac{b^{2} x}{a^{2}m}

 

CONJUGATE DIAMETERS:

Two diameters are said to be conjugate when each bisects all chords parallel to the other.

 

\\ {\text { If } y=m_{1} x \text { and } y=m_{2} x \text { be two conjugate diameters of an }} {\text { ellipse, then }}\\ m_{1} m_{2}=-\frac{b^{2}}{a^{2}}.

If PQ and RS be two conjugate diameters. Then the coordinates of the four extremities of two conjugate diameters are

\\ {P\equiv (a \cos \phi, b \sin \phi)} \\ {Q\equiv(-a \cos \phi,-b \sin \phi)} \\ {S\equiv(-a \sin \phi, b \cos \phi)} \\ {R\equiv(a \sin \phi,-b \cos \phi)}

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Locus of the mid-point of the chord of the ellipse \frac{x^2}{2}+y^2=1 is y=-\frac{x}{2m}                    ......(i)

Equation of a tangent to the ellipse x^2+\frac{y^2}{2}=1 in slope form is y = mx\pm\sqrt{m^2+2}                        .....(ii)

from eq (i) and eq (ii)

\\\Rightarrow y=\left(-\frac{x}{2 y}\right) x+\sqrt{\left(-\frac{x}{2 y}\right)^{2}+2}\\\Rightarrow \;\;\;\;\;(2y^2+x^2)^2={x^2+8y^2}

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rishi.raj

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