An ellipse having foci at (3, 3) and (-4, 4) | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the foci be $S_1(3, 3)$ and $S_2(-4, 4)$ and the point on the ellipse be $P(0, 0)$.By the definition of an ellipse,the sum of the distances fr

Vedclass · Accepted Answer

$\frac{5}{7}$

Vedclass · Answer

(C) Let the foci be $S_1(3, 3)$ and $S_2(-4, 4)$ and the point on the ellipse be $P(0, 0)$.By the definition of an ellipse,the sum of the distances from any point on the ellipse to the two foci is constant and equal to the length of the major axis,$2a$.$PS_1 = \sqrt{(3-0)^2 + (3-0)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$.$PS_2 = \sqrt{(-4-0)^2 + (4-0)^2} = \sqrt{16+16} = \sqrt{32} = 4\sqrt{2}$.Thus,$2a = PS_1 + PS_2 = 3\sqrt{2} + 4\sqrt{2} = 7\sqrt{2}$.The distance between the foci is $2ae = S_1S_2$.$S_1S_2 = \sqrt{(-4-3)^2 + (4-3)^2} = \sqrt{(-7)^2 + (1)^2} = \sqrt{49+1} = \sqrt{50} = 5\sqrt{2}$.Therefore,$2ae = 5\sqrt{2}$.Dividing the two equations: $e = \frac{2ae}{2a} = \frac{5\sqrt{2}}{7\sqrt{2}} = \frac{5}{7}$.

An ellipse having foci at $(3, 3)$ and $(-4, 4)$ and passing through the origin has eccentricity equal to

Similar Questions

The equation of the tangent to the ellipse $\frac{x^2}{4} + \frac{y^2}{12} = 1$ at the point $(1, 3)$ is:

The curve represented by $\frac{x^2}{12-\alpha} + \frac{y^2}{\alpha-10} = 1$ is

If $P \equiv (x, y)$,$F_1 \equiv (3, 0)$,$F_2 \equiv (-3, 0)$ and $16x^2 + 25y^2 = 400$,then $PF_1 + PF_2$ equals

If $4x - 3y - 5 = 0$ is a normal to the ellipse $3x^2 + 8y^2 = k$,then the equation of the tangent drawn to this ellipse at the point $(-2, m)$ $(m > 0)$ is

The tangent to the ellipse $9x^2 + 16y^2 = 288$ making equal intercepts on the co-ordinate axes intersects the $X$-axis and the $Y$-axis in the points $A$ and $B$ respectively. Then $\text{Area}(\triangle OAB) = $ (where $O$ is origin)

Vedclass Test Series

Exam Paper Generator

Online Exam Module