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A point on the ellipse x^2+3y^2=37 where...

A point on the ellipse `x^2+3y^2=37` where the normal is parallel to the line `6x-5y=2` is `(5,-2)` (b) (5, 2) (c) `(-5,2)` (d) `(-5,-2)`

A

(5,-2)

B

(5,2)

C

(-5,2)

D

(-5,-2)

Text Solution

Verified by Experts

The correct Answer is:
B
Differnetiating the equation of ellipse, `x^(2)+3y^(2)=37` w.r.t., x we get
`(dy)/(dx)=-(x)/(3y)`
The slop of the given line is 6/5 , which is normal to the ellipse
Hence, `3x//y=6//5 or 2x=5y`
Points in options (2) and (4) are satisfying the above equation adn that of ellipse.
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