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The equation of the normal to the ellips...

The equation of the normal to the ellipse at the point whose eccentric angle `theta=(pi)/(6)` is

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The correct Answer is:
`2[ax-sqrt(3) by] =sqrt(3)*(a^2 -b^2)`
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The equation of the normal to the ellipse x^(2)//16+y^(2)//9=1 at the point whose eccentric angle theta=pi//6 is

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Knowledge Check

  • The equation of the normal to the ellipse (x^(2))/(4)+(y^(2))/(2)=1 at the point whose eccentric angle is (pi)/(4) is

    A
    `x+sqrt(2)y=2sqrt(2)`
    B
    `sqrt(2)x-y=1`
    C
    `x-sqrt(2)y=0`
    D
    `sqrt(2)x+y=3`

  • The area of the parallelogram formed by the tangents at the points whose eccentric angles are theta, theta + pi/2, theta + pi, theta +(3pi)/2 on the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 is

    A
    ab
    B
    2ab
    C
    3ab
    D
    4ab

  • Locus of the point of intersection of the tangents at the points with eccentric angle theta and (pi)/(2)+theta is

    A
    `x^(2)+y^(2)=a^(2)`
    B
    `x^(2)+y^(2)=b^(2)`
    C
    `(x^(2))/(a^(2))+(y^(2))/(b^(2))=2`
    D
    `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1`

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