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The ellipse x^2/a^2+y^2/b^2=1,(a gt b) p...

The ellipse `x^2/a^2+y^2/b^2=1,(a gt b)` passes through `(2,3)` and have eccentricity equal to `1/2`. Then find the equation of normal to the ellipse at `(2,3)`.

A

2y-x=4

B

2x-y=1

C

3x-2y=0

D

3x-y=3

Text Solution

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The correct Answer is:
B
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Knowledge Check

  • If a tangent of slope 'm' at a point of m the ellipse (x^(2))/(a^(2))+ (y^(2))/(b^(2)) =1 passes through a b (2a, 0) and if 'e' denotes the eccentricity of the ellipse, then

    A
    `m^(2) + 2e^(2) =1 `
    B
    `3m^(2) +e^(2) =1`
    C
    `2m^(2) +e^(2)=1`
    D
    `m^(2) +e^(2) =`

  • Eccentricity of ellipse (x^(2))/(a^(2)) +(y^(2))/(b^(2))=1 if it passes through point (9, 5) and (12, 4) is

    A
    `sqrt(3//4)`
    B
    `sqrt(4//5)`
    C
    `sqrt(5//6)`
    D
    `sqrt(6//7)`

  • The equation of the ellipse passing through (2,1) and having e =1/2 is

    A
    `3x^2 + 4y^2 = 6`
    B
    `3x^2 + 5y^2 =17`
    C
    `5x^2 + 3y^2 = 23`
    D
    none

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