Question

Equation to the locus of the point which moves such that the sum of its distances from $(-4,3)$ and $(4,3)$ is $12$ is

• A. $\frac{x^2}{36}+\frac{(y-3{)}^2}{20}=1$
• B. $\frac{x^2}{20}+\frac{(y-3{)}^2}{36}=1$
• C. $\frac{(x-3{)}^2}{36}+\frac{y^2}{20}=1$
• D. $\frac{(x-1{)}^2}{36}+\frac{(y-3{)}^2}{20}=1$

Answer 1

Answer: (A)

$\frac{x^2}{36}+\frac{(y-3{)}^2}{20}=1$

We know that locus will be ellipse as sum of it's distances from two fixed points is constant.
Then focii of the ellipse are $(\pm 4,3)$

So, center of ellipse will be $(0,3)$ and constant distance is $12$ which is equal to $2a$(length of major axis)
Let $e$ be the eccentricity of ellipse

$2a$ and $2b$ be the lengths of major and minor axis respectively.
We know that distance between the focii is $2ae=8$, then

$e=\frac{2}{3}$ and also we know

$a^2(1-e^2)=b^2⟹b^2=20$
Now equation of ellipse be $\frac{x^2}{36}+\frac{{(y-3)}^2}{20}=1$