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Find the equation of the ellipse, the co...

Find the equation of the ellipse, the co-ordinates of whose foci are `(0, pm4)` and eccentricity `(4)/(5)`

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To find the equation of the ellipse given the coordinates of its foci and its eccentricity, we can follow these steps: ### Step 1: Identify the coordinates of the foci The coordinates of the foci are given as (0, ±4). This indicates that the foci are located on the y-axis. Therefore, we have: - \( c = 4 \) ### Step 2: Determine the major axis Since the foci are on the y-axis, the major axis of the ellipse is also along the y-axis. The standard form of the equation of an ellipse with a vertical major axis is: ...
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AAKASH INSTITUTE ENGLISH-CONIC SECTIONS-SECTION - J ( Aakash Challengers Questions )
  1. Find the equation of the ellipse, the co-ordinates of whose foci are ...

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  2. Find the angle between the two tangents from the origin to the circle ...

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  3. The area of the triangle formed by the tangent at (3, 4) to the circle...

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  4. If P(1), P(2), P(3) are the perimeters of the three circles, S(1) :...

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  5. If (1, a), (b, 2) are conjugate points with repect to the circle x^(2)...

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  6. Area of the equilateral triangle inscribed in the circle x^(2) + y^(2)...

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  7. A solid sphere of radius R/2 is cut out of a solid sphere of radius R ...

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  8. The range of parameter ' a ' for which the variable line y=2x+a lies b...

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  9. A planet of mass m moves along an ellipse around the sun (mass M) so t...

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  10. There are exactly two points on the ellipse x^2/a^2+y^2/b^2=1,whose di...

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  11. The line2px+ysqrt(1-p^(2))=1(abs(p)lt1) for different values of p, tou...

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  12. A point P moves such that the sum of the slopes of the normals drawn f...

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  13. A rectangular hyperbola whose centre is C is cut by any circle of radi...

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  14. Let P be a point on the hyperbola x^2-y^2=a^2, where a is a parameter,...

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  15. Tangents are drawn from the points on a tangent of the hyperbola x^2-y...

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  16. A tangent to the hyperbola (x^2)/(a^2)-(y^2)/(b^2)=1 cuts the ellipse ...

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  17. Let F(x) = (1+b^(2))x^(2) + 2bx + 1. The minimum value of F(x) is the ...

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