$\text{ The tangents are drawn from the points of a straight line }3x+4y=24\text{ to the curve }$

$x^2+\frac{y^2}{4}=1\text{. Then all the chords of contact passes through a fixed point }(a,b)\text{, then }2a+3b$

$\text{ Let the coordinates of a point lying on the straight line }3x+4y=24\text{ is }\left(t,\frac{24-3t}{4}\right)$

$\begin{array}{l}\text{ Equation of the chord of contact is }tx+\frac{y}{16}(24-3t)=1\\ \Rightarrow (24y-16)+t(16x-3y)=0\end{array}$

$\Rightarrow \text{ this line always passes through the fixed point which is the point of intersection of the }$

$\begin{array}{l}\text{ lines }24y-16=0\text{ and }16x-3y=0\\ \text{ fixed point }\equiv \left(\frac{1}{8},\frac{2}{3}\right),\text{ which lies on }16x-3y=0,9y^2=32x\text{ and }24x+24y=19.\end{array}$

$$\begin{gathered} \Rightarrow a=\frac{1}{8},b=\frac{2}{3} \\ \therefore 2a+3b=\frac{1}{4}+2=2.25 \end{gathered}$$