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Define hyperbola as a set of points deri...

Define hyperbola as a set of points derive its equation in the form ` (x^(2))/( a^(2))-( y^(2))/( b^(2)) =1 `

Text Solution

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The correct Answer is:
` Put c^(2)-a^(2)=b^(2)therefore x^(2)/a^(2)-y^(2)/b^(2) =1`
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Define ellipse and derive its equation in the form (x^(2))/(a^(2))+(y^(2))/(b^(2)) = 1(a gt b) .

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Knowledge Check

  • If the distance between the foci and the distance between the directrices of the hyperbola (x^(2))/(a^(2)) -(y^(2))/(b^(2)) =1 are in the ratio 3 : 2 then a : b is

    A
    `sqrt2: 1`
    B
    `1 : 2`
    C
    `sqrt3 : sqrt2`
    D
    `2 :1`

  • If the pair of lines b^(2) x^(2)-a^(2) y^(2)=0 are inclined at an angle theta then the eccentricity of the hyperbola (x^(2))/(a^(2))-(y^(2))/(b^(2))=1 is

    A
    1)`sec theta`
    B
    2)`cos theta`
    C
    3)`cos (theta)/(2)`
    D
    4)`sec ((theta)/(2))`

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