Find the equation of the ellipse with its cen | vedclass

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

Question

(B) The standard equation of an ellipse centered at the origin is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Since it passes through $(-3, 1)$ and $(2, -

Vedclass · Accepted Answer

$3x^2 + 5y^2 = 32$

Vedclass · Answer

(B) The standard equation of an ellipse centered at the origin is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Since it passes through $(-3, 1)$ and $(2, -2)$,we have:$\frac{9}{a^2} + \frac{1}{b^2} = 1$ (Equation $1$)$\frac{4}{a^2} + \frac{4}{b^2} = 1 \Rightarrow \frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{4}$ (Equation $2$)Subtracting Equation $2$ from Equation $1$:$(\frac{9}{a^2} - \frac{1}{a^2}) = 1 - \frac{1}{4}$ $\Rightarrow \frac{8}{a^2} = \frac{3}{4}$ $\Rightarrow a^2 = \frac{32}{3}$.Substituting $a^2$ into Equation $2$:$\frac{3}{32} + \frac{1}{b^2} = \frac{1}{4}$ $\Rightarrow \frac{1}{b^2} = \frac{8}{32} - \frac{3}{32} = \frac{5}{32}$ $\Rightarrow b^2 = \frac{32}{5}$.Since $a^2 = \frac{32}{3} \approx 10.67$ and $b^2 = \frac{32}{5} = 6.4$,the condition $a > b$ is satisfied.The equation is $\frac{x^2}{32/3} + \frac{y^2}{32/5} = 1$,which simplifies to $3x^2 + 5y^2 = 32$.

Find the equation of the ellipse with its center at the origin,passing through the points $(-3, 1)$ and $(2, -2)$,given that $a > b$.

Similar Questions

The eccentricity of the ellipse $y^{2}+4x^{2}-12x+6y+14=0$ is

The angle between the lines joining the foci of an ellipse to one particular extremity of the minor axis is $90^{\circ}$. The eccentricity of the ellipse is

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

Find the equation for the ellipse that satisfies the given conditions: Foci $(\pm 3, 0)$,$a = 4$.

Chords of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ are drawn through the positive end of the minor axis $(0, b)$. The locus of their midpoints lies on:

Vedclass Test Series

Exam Paper Generator

Online Exam Module