The equation 2x^2 + 3y^2 - 8x - 18y + 35 = k | vedclass

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

Question

(C) Given equation: $2x^2 + 3y^2 - 8x - 18y + 35 - k = 0$.Completing the square for $x$ and $y$ terms:$2(x^2 - 4x) + 3(y^2 - 6y) + 35 - k = 0$$2(x^2 -

Vedclass · Accepted Answer

$A$ point if $k = 0$

Vedclass · Answer

(C) Given equation: $2x^2 + 3y^2 - 8x - 18y + 35 - k = 0$.Completing the square for $x$ and $y$ terms:$2(x^2 - 4x) + 3(y^2 - 6y) + 35 - k = 0$$2(x^2 - 4x + 4) + 3(y^2 - 6y + 9) + 35 - k - 8 - 27 = 0$$2(x - 2)^2 + 3(y - 3)^2 = k$.Case $1$: If $k = 0$,the equation becomes $2(x - 2)^2 + 3(y - 3)^2 = 0$,which represents the point $(2, 3)$.Case $2$: If $k > 0$,the equation represents an ellipse.Case $3$: If $k Thus,the correct statement is that it represents a point if $k = 0$.

The equation $2x^2 + 3y^2 - 8x - 18y + 35 = k$ represents:

Similar Questions

The equations of the tangent and normal at point $(3, -2)$ of the ellipse $4x^2 + 9y^2 = 36$ are:

The equation of the normal at the point $(0, 3)$ of the ellipse $9x^2 + 5y^2 = 45$ is

Find the coordinates of the foci,the vertices,the length of the major axis,the minor axis,the eccentricity,and the length of the latus rectum of the ellipse $4x^{2} + 9y^{2} = 36$.

The centre of the ellipse $\frac{(x+y-3)^2}{9}+\frac{(x-y+1)^2}{16}=1$ is

The eccentricity of an ellipse,with its centre as origin,is $1/2$. If one of the directrices is $x=4$,then the equation of the ellipse is given by

Vedclass Test Series

Exam Paper Generator

Online Exam Module