The tangent to the parabola

To find intersection point of $x^{2}+y^{2}=5$ and $y^{2}=4 x$,

substitute $y^{2}=4 x$ in $x^{2}+y^{2}=5$, we get

$x^{2}+4 x-5=0 \Rightarrow x^{2}+5 x-x-5=0$

$\Rightarrow x(x+5)-1(x+5)=0$

$\therefore x=1,-5$

Intersection point in $1^{\text {st }}$ quadrant be $(1,2)$.

Now, equation of tangent to $y^{2}=4 x$ at $(1,2)$ is

$y \times 2=2(x+1) \Rightarrow y=x+1$

$\Rightarrow x-y=0$ .....$\ldots(\mathrm{i})$

Hence, $\left(\frac{3}{4}, \frac{7}{4}\right)$ lies on (i)