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Q.
If the line $3x - 4y + 5 = 0$ is a tangent to the parabola $y^2 = 4ax,$ then $a$ is equal to
Conic Sections
Solution:
The given line is $y= \frac{3}{4}x+\frac{5}{4} = mx +c$
where $m=\frac{3}{4}, c $
$ = \frac{5}{4} $
$ y= mx+c $ touches $y^{2}= 4ax$ if $c= \frac{a}{m} $
$ \therefore $ the given line touches $y^{2}= 4ax$
if $\frac{5}{4}= \frac{a}{{3}/{4}} $
$ \Rightarrow a=\frac{5}{4}\times\frac{3}{4}$
$ = \frac{15}{16}$