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On the ellipse 4x^2+9y^2=1, the points a...

On the ellipse `4x^2+9y^2=1,` the points at which the tangents are parallel to the line `8x=9y` are (a)`(2/5,1/5)` (b) `(-2/5,1/5)` (c)`(-2/5,-1/5)` (d) `(2/5,-1/5)`

A

`((2)/(5), (1)/(5))`

B

`(-(2)/(5), (1)/(5))`

C

`(-(2)/(5), - (1)/(5))`

D

`((2)/(5), - (1)/(5))`

Text Solution

Verified by Experts

The correct Answer is:
B, D
Given, ` 4x ^(2 ) + 9y^(2) = 1 " " `… (i)
On differentiating w.r.t. x, we get
`" " 8x + 18y (dy)/(dx) =0`
`rArr " " (dy)/(dx) =- ( 8x)/( 18y ) =- ( 4x )/( 9y)`
The tangent at point `(h, k)` will be parallel to ` 8x = 9y`, then
`" " - ( 4h)/( 9k) = (8)/(9)`
`rArr " " h = - 2k `
Point `(h, k)` also lies on the ellipse.
`therefore " " 4h^(2) + 9k^(2) = 1" " `... (ii)
On putting value of h in Eq. (ii), we get
`" " 4 (-2 k)^(2) + 9k ^(2) = 1 `
`rArr " " 16 k ^(2) = 9 k^(2) = 1 `
`rArr " " 25 k^(2) = 1 `
`rArr " " k ^(2) = (1)/( 25)`
`rArr " " k = pm (1)/(5)`
Thus, the point, where the tangents are parallel to `8x = 9y ` are `(-(2)/(5), (1)/(5)) and ((2)/(5), - (1)/(5))`.
Therefore, options (b) and (d) are the answers.
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