For the ellipse 4x^2 + y^2 - 8x + 2y + 1 = 0, | vedclass

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

Question

(D) Given equation: $4x^2 + y^2 - 8x + 2y + 1 = 0$ Completing the square: $4(x^2 - 2x) + (y^2 + 2y) + 1 = 0$ $4(x-1)^2 - 4 + (y+1)^2 - 1 + 1 = 0$ $4(x

Vedclass · Accepted Answer

$(1, -1-\sqrt{3}), \sqrt{3}y+\sqrt{3}+4=0$

Vedclass · Answer

(D) Given equation: $4x^2 + y^2 - 8x + 2y + 1 = 0$ Completing the square: $4(x^2 - 2x) + (y^2 + 2y) + 1 = 0$ $4(x-1)^2 - 4 + (y+1)^2 - 1 + 1 = 0$ $4(x-1)^2 + (y+1)^2 = 4$ Dividing by $4$: $\frac{(x-1)^2}{1} + \frac{(y+1)^2}{4} = 1$ Here,$a^2 = 1$ and $b^2 = 4$. Since $b^2 > a^2$,the major axis is vertical. Eccentricity $e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$. Center $(h, k) = (1, -1)$. Foci are given by $(h, k \pm be) = (1, -1 \pm 2 \cdot \frac{\sqrt{3}}{2}) = (1, -1 \pm \sqrt{3})$. Directrices are given by $y = k \pm \frac{b}{e} = -1 \pm \frac{2}{\sqrt{3}/2} = -1 \pm \frac{4}{\sqrt{3}}$. For the focus $(1, -1-\sqrt{3})$,the corresponding directrix is $y = -1 - \frac{4}{\sqrt{3}}$ $\Rightarrow \sqrt{3}y = -\sqrt{3} - 4$ $\Rightarrow \sqrt{3}y + \sqrt{3} + 4 = 0$.

For the ellipse $4x^2 + y^2 - 8x + 2y + 1 = 0$,the focus and the equation of the directrix are respectively

Similar Questions

If the points of intersection of the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1$ and the circle $x^{2}+y^{2}=4b$ (where $b > 4$) lie on the curve $y^{2}=3x^{2}$,then $b$ is equal to:

The position of the point $(4, -3)$ with respect to the ellipse $2x^2 + 5y^2 = 20$ is

The value of $\lambda$,for which the line $2x - \frac{8}{3}\lambda y = -3$ is a normal to the conic $x^2 + \frac{y^2}{4} = 1$ is

The equation of an ellipse whose focus is $(-1, 1)$,whose directrix is $x - y + 3 = 0$,and whose eccentricity is $e = \frac{1}{2}$,is given by

The sides of the rectangle of greatest area that can be inscribed in the ellipse $x^2+4y^2=64$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module