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The tangents from (1, 2sqrt2) to the hyp...

The tangents from `(1, 2sqrt2)` to the hyperbola `16x^2-25y^2 = 400` include between them an angle equal to:

Text Solution

Verified by Experts

The correct Answer is:
`y=-3xpmsqrt(209)`
We have hyperbola `(x^(2))/(25)-(y^(2))/(16)=1`.
The slope of given line `x-3y=4` is 1/3.
So, the slope of tangent is `-3.`
Therefore, equations of tangests are
`y=-3xpmsqrt(25xx9-16)`
`"or "y=-3x pm sqrt(209)`
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Knowledge Check

  • The vertices of the hyperbola 9x^2 - 16y^2 = 144

    A
    `(+- 5, 0 )`
    B
    `(+- 4, 0)`
    C
    `(+- 5/4 , 0)`
    D
    `(+- 3/4 , 0)`

  • The tangent to the hyperbola 3x^(2)-y^(2)=3 parallel to 2x-y+4=0 is:

    A
    `2x-y-3=0`
    B
    `2x-y+1=0`
    C
    `2x+ypm1=0`
    D
    `2x-y-1=0`

  • The eccentricity of the ellipse 16x^2 + 25y^2 = 400 is

    A
    `3/5`
    B
    `3/4`
    C
    `4/5`
    D
    `9/25`

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