Question

If the focus is $(\alpha ,\beta )$ & the directrix is $ax+by+c=0$ then the equation of conic whose eccentricity $=e$ is given by ${(x-\alpha )}^2+{(y-\beta )}^2=e^2\frac{{(ax+by+c)}^2}{a^2+b^2}$. If $e=1$ then conic is called parabola, for $e<1$ (conic is an ellipse) and for $e>1,$ conic is a hyperbola.

Now consider the conic

$169\{{(x-1)}^2+{(y-3)}^2\}={(5x-12y+17)}^2$ ......$(\ast )$

On the basis of above information answer the following question:

The equation of axis of the conic $(\ast )$ is

• A. $12x+5y+20=0$
• B. $12x-5y+17=0$
• C. $12x+5y+17=0$
• D. $12x+5y-27=0$

Answer 1

The correct option is D $12x+5y-27=0$

A conic is the locus of a point'$P$' which moves in such a way that
its distances from a fixed point'$S$' always bears a constant ratio
to its distances from a fIxed straight line.
The fixed point '$S$'
is called focus. The fixed straight line is called directrix' & the
constant ratio' is known as eccentricity denoted by $e$.
$\therefore e=PS/PM$
Now $169\{{(x-1)}^2+{(y-3)}^2\}={(5x-12y+17)}^2$
$\Rightarrow \{{(x-1)}^2+{(y-3)}^2\}=\frac{{(5x-12y+17)}^2}{5^2+(-12{)}^2}={\left(\frac{5x-12y+17}{13}\right)}^2$
$\therefore {(x-1)}^2+{(y-3)}^2=e^2{\left(\frac{5x-12y+17}{13}\right)}^2$ where $e=1$
Any line passing through focus $(1,3)$ and perpendicular to $5x-12y+17=0$ is the axis of conic (*). Now any line
$\perp$er to $5x-12y+17=0$ is given by
$12x+5y+\lambda =0$ but it passes through focus
$\therefore \lambda =-12\left(1\right)-5\left(3\right)=-27$
$\therefore$ equation of axis is $12x+5y-27=0$