mirror equation...hate them!!

You haven't mentioned whether it is a concave or convex mirror. So I am taking cases.

Using the mirror equation,

$\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$
where:
$u$ - object distance
$v$ - image distance
$f$ - focal length of the mirror
(NOTE: The equation is formulated defining the pole as origin and the quantities $u,v\ \mbox{and}\ f$ denote the x-coordinates)

CASE I(For concave mirror):
We have,
$f$ = - 26.6 cm (Considering focus towards left of pole as it is a concave mirror)

So using mirror equation,
$\frac{1}{u} + \frac{1}{v} = -\frac{1}{26.6}$

$\frac{u + v}{uv} = -\frac{1}{26.6}$ --------- ■

We know that,

Magnification = $\frac{\mbox{Height of image}}{\mbox{Height of object}}$ = $-\frac{v}{u}$

(The equation is formulated taking proper signs)

Thus,
$- 0.35 = -\frac{v}{u}$

$\implies v = 0.35u$ -------- ●

Using the above relation in ■,
$\frac{u(0.35 + 1)}{0.35u^2} = -\frac{1}{26.6}$

$\implies (1.35\times 26.6)u = -0.35u^2$

$\implies 35.91u + 0.35u^2 = 0$

$\implies u(35.91 + 0.35u) = 0$

$\implies u = 0$ (rejected as image cannot be formed at the pole)
or $u = -\frac{35.91}{0.35}\ \mbox{cm} = - 102.6\ \mbox{cm}$

Using ●,
$v = 0.35\times -102.6 = -35.91\ \mbox{cm}$

Thus, the image is formed at a distance of $\boxed{35.91\ \mbox{cm}}$ towards the left of the concave mirror. (The negative sign implies negative x-axis which is towards the left of the origin i.e the pole)

Similarly,
CASE II(For Convex mirror):

We have,
$f$ = + 26.6 cm [Considering focus to the right of the pole]

Repeating the rest of the process,
$v = +35.91\ \mbox{cm}$

Thus, the image is formed at a distance of $\boxed{35.91\ \mbox{cm}}$ towards the right of the convex mirror.
(The positive sign implies positive x-axis which is towards the right of the origin i.e. the pole)