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. $3x^2+4y^2+42x+120=0$
• C. $3x^2+4y^2+42x–120=0$
• D. $3x^2+4y^2–42x–120=0$

Answer 1

The correct option is A

$3x^2+4y^2–42x+120=0$

Explanation for the correct option

Finding the equation of the ellipse

Centre of ellipse is the mid point of the vertices.

Calculating the center of the ellipse,

$$\begin{gathered} =\left(\frac{4+10}{2},\frac{0+0}{2}\right) \\ =(7,0) \end{gathered}$$

Now,

$$\begin{gathered} 2a=6\left[2timesmajoraxisisdis\tan cebetweentwopoints\right] \\ a=3 \end{gathered}$$

$$\begin{gathered} e=\frac{1}{2} \\ {b}^2=a^2(1-e^2)\left[Relationbetweenmajoraxisandminoraxis\right] \\ =3^2(1-{\frac{1}{4}}^{}) \\ =9\left(\frac{3}{4}\right) \\ =\frac{27}{4} \end{gathered}$$

Hence, the equation of the ellipse is,

$\begin{array}{rcl}\frac{{(x-7)}^2}{9}+\frac{y^2}{{\displaystyle \frac{27}{4}}}& =& 1\\ \frac{{(x-7)}^2}{9}+\frac{4y^2}{{\displaystyle 27}}& =& 1\\ 3(x^2-14x+49)+4y^2& =& 27\\ 3x^2-42x+147+4y^2-27& =& 0\\ 3x^2-42x+120+4y^2& =& 0\end{array}$

Hence, Option(A)is the correct answer