The equation of the ellipse with its focus at | vedclass

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

Question

(D) Given,focus $S = (6, 2)$,centre $C = (1, 2) = (h, k)$,and the ellipse passes through $P = (4, 6)$.The standard equation of the ellipse is $\frac{(

Vedclass · Accepted Answer

$\frac{(x-1)^2}{45}+\frac{(y-2)^2}{20}=1$

Vedclass · Answer

(D) Given,focus $S = (6, 2)$,centre $C = (1, 2) = (h, k)$,and the ellipse passes through $P = (4, 6)$.The standard equation of the ellipse is $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$.Substituting the centre $(1, 2)$,we get $\frac{(x-1)^2}{a^2} + \frac{(y-2)^2}{b^2} = 1$ ... $(i)$.Since the ellipse passes through $P(4, 6)$,we have $\frac{(4-1)^2}{a^2} + \frac{(6-2)^2}{b^2} = 1$,which simplifies to $\frac{9}{a^2} + \frac{16}{b^2} = 1$ ... (ii).The distance from the centre to the focus is $ae = 6 - 1 = 5$,so $a^2e^2 = 25$.Using the relation $b^2 = a^2(1 - e^2) = a^2 - a^2e^2$,we get $b^2 = a^2 - 25$,or $a^2 = b^2 + 25$ ... (iii).Substituting $a^2$ into (ii): $\frac{9}{b^2+25} + \frac{16}{b^2} = 1$.$9b^2 + 16(b^2 + 25) = b^2(b^2 + 25) \implies 25b^2 + 400 = b^4 + 25b^2 \implies b^4 = 400 \implies b^2 = 20$.From (iii),$a^2 = 20 + 25 = 45$.Thus,the equation is $\frac{(x-1)^2}{45} + \frac{(y-2)^2}{20} = 1$.

The equation of the ellipse with its focus at $(6,2)$,centre at $(1,2)$,and which passes through the point $(4,6)$ is

Similar Questions

Find the coordinates of the foci,the vertices,the length of the major axis,the minor axis,the eccentricity,and the latus rectum of the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$.

The acute angle between the pair of tangents drawn to the ellipse $2x^{2} + 3y^{2} = 5$ from the point $(1, 3)$ is:

The equation of the locus of the foot of the perpendicular drawn from the centre of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ to any tangent of the ellipse is

For $\alpha$ belonging to an interval of length $\beta$,suppose $(\alpha, -\alpha)$ is an interior point of the ellipse $4x^2 + 5y^2 = 1$. Then,$(6\beta - 4)^{201} + 201 = $

From the point $C(0, \lambda)$,two tangents are drawn to the ellipse $x^2 + 2y^2 = 4$,cutting the major axis at $A$ and $B$. If the area of $\Delta ABC$ is minimum,then the value of $\lambda$ is-

Vedclass Test Series

Exam Paper Generator

Online Exam Module