Question

The equation of an ellipse whose eccentricity is $\frac{1}{2}$ and the vertices are $(4,0)$ and $(10,0)$ is

• A. $3x^2+4y^2-42x+120=0$
• B. $3x^2+4y^2-40x+130=0$
• C. $3x^2+4y^2-38x+140=0$
• D. $3x^2+4y^2-36x+150=0$

Answer 1

Answer: (A)

$3x^2+4y^2-42x+120=0$

Centre of ellipse is the mid point of the vertices.

Centre$=(\frac{4+10}{2},\frac{0+0}{2})=(7,0)$

$2a=6$

$\Rightarrow a=3$

$e=\frac{1}{2}$

$b^2=a^2(1-e^2)$

$=3^2(1-\frac{1}{4})=9\left(\frac{3}{4}\right)=\frac{27}{4}$

Hence, the equation of the ellipse is

$\frac{(x-7{)}^2}{9}+\frac{y^2}{\frac{27}{4}}=1$

(or) $\frac{(x-7{)}^2}{9}+\frac{4y^2}{27}=1$

$3(x^2-14x+49)+4y^2=27$

$3x^2+4y^2-42x+147-27=0$

$3x^2+4y^2-42x+120=0$.