The limiting value of expression $\frac{4x^2+2y^2-6xy}{6x^2+√2y-8xy}$ is $A$ as point $(x,y)$ on curve $x^2+y^2=1$ approaches the position $(\frac1{√2},\frac1{√2})$ where $A$ is such that $(5A,0)$ is a point as focus of parabola $S$ having axis parallel to x-axis, vertex at origin. The two common tangents can be drawn to both circle and parabola from an external point; find the coordinates of that point. Also find the locus of midpoints of chords of parabola, which subtend a right angle at vertex of parabola.
My Attempt:
At first I thought there is a typo in the denominator and it should be $√2y^2$ but the expression as written is of the form $0/0$, so, then I thought of removing the factors causing $0/0$ form.
Numerator can be factorized as $2(2x-y)(x-y)$. Not able to factorize the denominator.
Then I replaced $x$ by $\cos\theta$ and $y$ by $\sin\theta$, and got
$\frac{(cos\theta-\sin\theta)(3+\cos\theta+\sin\theta)}{3+3\cos2\theta+√2\sin\theta-4\sin2\theta}$,
which is still a $0/0$ form. Don't know how to solve it.
Add and subtract $2y^2$ in the denominator
$$6x^2+\sqrt{2}y-8xy+2y^2-2y^2 = 2(3x^2-4xy+y^2)+\sqrt{2}y-2y^2$$
$$ = 2(3x-y)(x-y)+\sqrt{2}y(1-\sqrt{2}y)$$
Combined with the factoring you already recognized, this means we can rewrite your limmand as
$$\frac{\sqrt{2}(2x-y)}{\sqrt{2}(3x-y)+\frac{y(1-\sqrt{2}y)}{(x-y)}} \longrightarrow \frac{1}{2+L}\equiv A$$
as we approach $\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$. Now $L$ doesn't exist in general as the limit can grow unboundedly along certain paths, for example along $x=y+(1-\sqrt{2}y)^2$.
But we are being told to restrict the calculation of $L$ to one path (along the unit circle), which we will denote $L_{S^1}$. Plug in the path $x = \sqrt{1-y^2}$:
$$\require{cancel}L_{S^1} = \lim_{y\to\frac{1}{\sqrt{2}}}\frac{y(1-\sqrt{2}y)}{\sqrt{1-y^2}-y} = \lim_{y\to\frac{1}{\sqrt{2}}}\frac{y(\sqrt{1-y^2}+y)}{1+\sqrt{2}y}\cdot\frac{\cancel{(1-\sqrt{2}y)}}{\cancel{(1-\sqrt{2}y)}} = \frac{1}{2}$$
Therefore your limit is $\boxed{A = \frac{2}{5}}$