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Derive the equation of the ellipse in th...

Derive the equation of the ellipse in the form `(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) .

If the area of the auxillary circle of the ellipse (x^(2))/(a^(2)) + (y^(2))/(b^(2)) = 1 (a gt b) is twice the area of the ellipse, then the eccentricity of the ellipse is

The equation of the normal to the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 at the end of the latus rectum in the first quadrant is

If the area of the auxilliary circle of the ellipse (x^(2))/(a^(2)) + (y^(2))/(b^(2)) =1 (a gt b) is twice the area of the ellipse, then the eccentricity of the ellipse is

AOB is the positive quadrant of the ellipse (X^(2))/(a^(2)) + (y^(2))/(b^(2)) =1 , where OA =a, OB =b. The area between the arc AB and the chord AB of the ellipsed is

If e is the eccentricity of the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 , (altb) then

Find the equation of hyperbola in the form (x^2)/(a^2)-(y^2)/(b^2)=1 , given that length of transverse axis = 10 and eccentricity = 2.

The locus of the middle point of the portion of a tangent to the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 included between the axes is the curve

OSWAAL PUBLICATION-SAMPLE PAPER 4-EXERCISE
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