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If the normal to the ellipse 3x^(2)+...

If the normal to the ellipse `3x^(2)+4y^(2)=12` at a point P on it is parallel to the line , `2x-y=4` and the tangent to the ellipse at P passes through Q (4,4) then Pq is equal to

A

`(sqrt(61))/(2)`

B

`(sqrt(157))/(2)`

C

`(sqrt(221))/(2)`

D

`(5sqrt5)/(2)`

Text Solution

Verified by Experts

The correct Answer is:
D
`3x^(2)+4y^(2)=12 rArr 8y(dy)/(dx)=-6x`
`(dy)/(dx)=-(6)/(8)(x)/(y)" "rArr" slope of normal"-(dx)/(dy)=(4y)/(3x)=-2" (given)"`
`2y=-3x`
Also, `x,y,` lies on `3x^(2)+4y^(2)=12`
`((2y)/(3))^(2)+4y^(2)=12," "4y^(2)+12y^(2)=36 rArr x = pm1(pm1, pm(3)/(2))`
i.e. `p(-1, (3)/(2)) or (4,4 )` (A little consideration shown that P must lie in `2^("nd")` quadrant)
`rArr" "sqrt(5^(2)+((5)/(2))^(2))=sqrt(25+(25)/(4))=(5sqrt5)/(2)`
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