The maximum value of the sum of the intercepts made by any tangent to the curve $\left(a{\sin }^2\theta ,2a\sin \theta \right)$ with the axes is

 The curve is $\left(a{\sin }^2\theta ,2a\sin \theta \right)\Rightarrow x=a{\sin }^2\theta ,y=2a\sin \theta$ 

Now, the equation of any tangent to the curve is,

$$\begin{gathered} \frac{y-2a\sin \theta }{x-a{\sin }^2\theta } =\frac{1}{\sin \left(\theta \right)} \\ \Rightarrow y\sin \left(\theta \right) =x+a{\sin }^2\theta \end{gathered}$$

$\Rightarrow \frac{x}{-a{\sin }^2\theta }\frac{y}{a\sin \theta }=1$

$\Rightarrow$ Sum of intercepts $=a\left({\sin }^2\theta \sin \theta \right)$

$=a\cdot \left\{{\left(\sin \theta +\frac{1}{2}\right)}^2 -\frac{1}{4}\right\}$

Which is maximum, when $\sin \theta =1$

i.e. $($ Sum of intercept) max $=2a$

hence ,the maximum value of  the intercept made by the tangent to curve is $2a$