Question

The equation $(13x-1{)}^2+(13y-1{)}^2=k(5x-12y+1{)}^2$ will represent a parabola if

• A. $k=2$
• B. $k=81$
• C. $k=169$
• D. $k=1$

Answer 1

Answer: (D)

$k=1$

We know that the general equation of the ellipse is

${(x-h)}^2+{(y-k)}^2=e\frac{{(lh+mk+n)}^2}{l^2+m^2}$ ................$\left(1\right)$

${(13x-1)}^2+{(13y-1)}^2=k{(5x-12y+1)}^2$

$\Rightarrow {13}^2{(x-\frac{1}{13})}^2+{13}^2{(y-\frac{1}{13})}^2=k{(5x-12y+1)}^2$

$\Rightarrow 169[{(x-\frac{1}{13})}^2+{(y-\frac{1}{13})}^2]=k{(5x-12y+1)}^2$

$\Rightarrow {(x-\frac{1}{13})}^2+{(y-\frac{1}{13})}^2=\frac{k}{169}{(5x-12y+1)}^2$ ...........$\left(2\right)$

A parabola has its eccentricity $e=1$

Comparing equations $\left(1\right)$ and $\left(2\right)$ we get

$\Rightarrow k=1$

$\therefore k=1$