Centre of Ellipse

Q. 34. The length of normal (terminated by major axis) at a point of the ellipse x2/a2+y2/b2=1

Q.

Find the equation of the ellipse for which $e=\frac{4}{5}$ and whose vertices are $(0,\pm 10).$

Q.

An ellipse with centre at (0, 0) cuts x axis at (3, 0) and (-3, 0). If its eccentricity is $\frac{1}{2}$ then the length of its semiminor axis is

1. $2\sqrt{3}$

2. $\sqrt{5}$

3. $3\sqrt{2}$

4. $\frac{3\sqrt{3}}{2}$

Q. The eccentricity of an ellipse, with its centre at the origin, is $\frac{1}{2}$. If one of the directrices is x = 4, then the equation of the ellipse is

1. 
2. 
3. 
4. 

Q.

Find the equation for the ellipse that satisfies the given conditions,
$\text{Vertices}(\pm 6,0),\text{foci}(\pm 4,0)$

Q. A normal to $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ meets the axes in L and M. The perpendiculars to the axes through L and M intersect at P. Then the equation to the locus of P is

1. $a^2{x}^2-b^2{y}^2=(a^2+b^2{)}^2$
2. $a^2{x}^2+b^2{y}^2=(a^2+b^2{)}^2$
3. $b^2{x}^2-a^2{y}^2=(a^2-b^2{)}^2$
4. $a^2{x}^2+b^2{y}^2=(a^2+b^2{)}^2$

Q.

Centre of the Ellipse $4(x-2y+1{)}^2+9(2x+y+2{)}^2$ = 5 is

1. (-5, 2)

2. (-2, 2)

3. (-1, 0)

4. (1, 5)