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Convert the following equation of ellips...

Convert the following equation of ellipse into standard from .
(i) `16x^(2)+9y^(2)=144`
(ii) `9x^(2)+25y^(2)=225`

Text Solution

Verified by Experts

The correct Answer is:
(i) `(x^(2))/(3^(2))+(y^(2))/(5^(2))=1`, (ii) `(x^(2))/(5^(2))+(5^(2))/(3^(2))=1`
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25x^(2)-9y^(2)=225

Write the equation of ellipse 25x^(2)+9y^(2)=225 in standard form.

Knowledge Check

  • Equation of a common tangent to x^(2)+y^(2)=16" and "9x^(2)+25y^(2)=225 is :

    A
    `3x+2y=5`
    B
    `sqrt(7x)+3y=16`
    C
    `x+sqrt(7)y=6`
    D
    `sqrt(7)x-sqrt(11)y=2`

  • A common tangent to 9x^(2)-16y^(2)=144and x^(2)+y^(2)=9 is

    A
    `y=(3)/(sqrt(7))x+(15)/(sqrt(7))`
    B
    `y=3sqrt((2)/(7))x+(15)/(sqrt(7))`
    C
    `y=2sqrt((3)/(7))x+15sqrt(7)`
    D
    none of these

  • A common tangent to 9x^(2) - 16y^(2) = 144 and x^(2) + y^(2) = 9 is

    A
    `y sqrt(7) = sqrt(2)x + 15`
    B
    `ysqrt(7) = 3sqrt(2)x + 15`
    C
    `y = 3sqrt(2)x + 15`
    D
    `ysqrt(7) = 3x + 15`

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    For the following ellipse, find the equation of directrix (i) 9x^(2) + 4y^(2) = 36 (ii) 16x^(2) + 25y^(2) = 400 .

    The equation of the tangent to the hyperbola 16x^(2)-9y^(2)=144 at (5,8//3) , is

    Find the lengths of transverse and conjugate axes, co-ordinates of foci, vertices and the eccentricity for the following hyperbolas : (i) 16x^(2) - 9y^(2) = 144 (ii) 2x^(2) - 3y^(2) - 6 =0 (iii) 3x^(2) - 2y^(2) = 1

    A common tangent to 9x^(2)-16y^(2)=144 and x^(2)+y^(2)=9, is

    Equation of het hyperbola whose vertices are (+-3,0) and foci at (+-5,0) is 16x^(2)-9y^(2)=144 b.9x^(2)-16y^(2)=144c25x^(2)-9y^(2)=225d.9x^(2)-25y^(2)=81

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