Find the equation of the ellipse whose focus is $(-1, 1)$,eccentricity is $1/2$,and directrix is $x - y + 3 = 0$.

  • A
    $7x^{2} + 14y^{2} + 4xy + 12x - 10y + 12 = 0$
  • B
    $7x^{2} + 7y^{2} + 2xy + 10x - 10y + 7 = 0$
  • C
    $14x^{2} + 7y^{2} + 4xy + 12x - 10y + 14 = 0$
  • D
    None of these

Explore More

Similar Questions

If the points of intersection of two distinct conics $x^2+y^2=4b$ and $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ lie on the curve $y^2=3x^2$,then $3\sqrt{3}$ times the area of the rectangle formed by the intersection points is............................

If $e$ and $e^{\prime}$ are the eccentricities of the ellipse $5x^2 + 9y^2 = 45$ and the hyperbola $5x^2 - 4y^2 = 45$ respectively,then $ee^{\prime}$ is equal to

Consider two families of curves $y^2=4ax$ ($a$ is a parameter) and $x^2+\frac{y^2}{2}=c^2$ ($c$ is a parameter). If one curve from each family is chosen,then the angle between those two curves is

Let the eccentricity of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ be the reciprocal of the eccentricity of the hyperbola $2x^2 - 2y^2 = 1$. If the ellipse intersects the hyperbola at right angles,then the square of the length of the latus-rectum of the ellipse is $................$.

The slopes of the common tangents to the parabola $(x - 1)^2 = 4(y - 2)$ and the ellipse $\frac{(x - 1)^2}{1} + \frac{(y - 2)^2}{2} = 1$ are $m_1$ and $m_2$. Then,$m_1^2 + m_2^2$ is equal to:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo