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Find the locus of the middle points of all chords of `(x^2)/4+(y^2)/9=1` which are at a distance of 2 units from the vertex of parabola `y^2=-8a xdot`

Text Solution

Verified by Experts

The correct Answer is:
`4((x^(2))/(16)+(y^(2))/(81))=((x^(2))/(4)+(y^(2))/(9))`
The equation of the chord of ellipse which gets bisected at P(h,k) is
`(hx)/(4)+(ky)/(9)=(h^(2))/(4)+(k^(2))/(9)" "(1)`
Ist distance from the rogin (0,0) is 2. Therefore.
`(|0+0-((h^(2))/(4)-(k^(2))/(9))|)/(sqrt((h^(2))/(16)+k^(2)/(81)))=2`
Therefore, the locus is
`4((x^(2))/(16)+(y^(2))/(81))=((x^(2))/(4)+(y^(2))/(9))^(2)`
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