Find the equation for the ellipse that satisf | vedclass

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

Question

(A) Since the major axis is on the $x-$ axis,the equation of the ellipse is of the form $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ $(1)$.Given th

Vedclass · Accepted Answer

$\frac{x^{2}}{52} + \frac{y^{2}}{13} = 1$

Vedclass · Answer

(A) Since the major axis is on the $x-$ axis,the equation of the ellipse is of the form $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ $(1)$.Given that the ellipse passes through $(4, 3)$ and $(6, 2)$,we have:$\frac{16}{a^{2}} + \frac{9}{b^{2}} = 1$ $(2)$$\frac{36}{a^{2}} + \frac{4}{b^{2}} = 1$ $(3)$Let $u = \frac{1}{a^{2}}$ and $v = \frac{1}{b^{2}}$. The equations become:$16u + 9v = 1$$36u + 4v = 1$Multiplying the first by $4$ and the second by $9$:$64u + 36v = 4$$324u + 36v = 9$Subtracting the equations: $260u = 5 \implies u = \frac{5}{260} = \frac{1}{52}$.Thus,$a^{2} = 52$.Substituting $u$ into $36u + 4v = 1$:$36(\frac{1}{52}) + 4v = 1 \implies \frac{9}{13} + 4v = 1 \implies 4v = 1 - \frac{9}{13} = \frac{4}{13} \implies v = \frac{1}{13}$.Thus,$b^{2} = 13$.The equation is $\frac{x^{2}}{52} + \frac{y^{2}}{13} = 1$.

Find the equation for the ellipse that satisfies the given conditions: Major axis on the $x-$ axis and passes through the points $(4, 3)$ and $(6, 2)$.

Similar Questions

If $S$ and $S^{\prime}$ are the foci of an ellipse,$B$ is one end of the minor axis and $\angle SBS^{\prime} = 90^{\circ}$,then the eccentricity of that ellipse is

The conic represented by $x = 2(\cos t + \sin t), y = 5(\cos t - \sin t)$ is .....

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

The eccentricity of an ellipse,with its centre at the origin,is $\frac{1}{2}$. If one of the directrices is $x = 4$,then the equation of the ellipse is

The least intercept made by a tangent to the ellipse $\frac{x^2}{64}+\frac{y^2}{49}=1$ with the coordinate axes is

Vedclass Test Series

Exam Paper Generator

Online Exam Module