If the tangent at the point (1, 2) on the ellipse $3{\mathrm{x}}^2+4{\mathrm{y}}^2=19$ is also a tangent to the parabola ${\mathrm{y}}^2-\mathrm{kx}=0,$ then k =

Given ellipse $\mathrm{S}\equiv 3{\mathrm{x}}^2+4{\mathrm{y}}^2-19=0$

Given point P=(1, 2)

Now equation of tangent at P is ${\mathrm{S}}_1=0$

$$\begin{gathered} \Rightarrow 3\mathrm{x}\left(1\right)+4\mathrm{y}\left(2\right)-19=0 \\ \Rightarrow 3\mathrm{x}+8\mathrm{y}-19=0 \\ \Rightarrow \mathrm{y}=\frac{-3}{8}\mathrm{x}+\frac{19}{8} \end{gathered}$$

Given parabola ${\mathrm{y}}^2=\mathrm{k} \mathrm{x}$

where $4\mathrm{a}=\mathrm{k} \Rightarrow \mathrm{a}=\frac{\mathrm{k}}{4}$

since (1) is also tangent to the parabola then $c=\frac{\mathrm{a}}{\mathrm{m}}$

$$\begin{gathered} \Rightarrow \frac{19}{8}=\frac{{\displaystyle \frac{\mathrm{k}}{4}}}{{\displaystyle \frac{-3}{8}}} \\ \Rightarrow \frac{19}{8}=\frac{-8\mathrm{k}}{12} \\ \Rightarrow -64\mathrm{k}=19\times 12 \\ \Rightarrow \mathrm{k}=\frac{-19\times 12}{64}=\frac{-57}{16} \end{gathered}$$