Answer: 7

Solution:

Formula: Equation of Tangent( Cartesian form ), Distance Formula

Given curve,

$ \begin{aligned} & 9 x^2+16 y^2=144 \\ & \Rightarrow \frac{x^2}{16}+\frac{y^2}{9}=1 \\ & \Rightarrow \frac{x^2}{4^2}+\frac{y^2}{3^2}=1 \\ & \therefore a=4 \text { and } b=3 \end{aligned} $

So, general point on the ellipse is $ (4 \cos \theta, 3 \sin \theta) $

We know, Equation of tangent to a given ellipse at its point $(a \cos \theta, b \sin \theta)$ is

$ \frac{x \cos \theta}{a}+\frac{y \sin \theta}{b}=1 $

$\therefore$ Here equation of tangent at point $(4 \cos \theta, 3 \sin \theta)$ is $ \frac{x \cos \theta}{4}+\frac{y \sin \theta}{3}=1 $

When this tangent cut’s $x$ axis then $y=0$. $ \begin{aligned} & \therefore \frac{x \cos \theta}{4}+0=1 \\ & \Rightarrow x=4 \sec \theta \end{aligned} $

$\therefore$ Point of intersection at $x$ axis is $A(4 \sec \theta, 0)$.

When this tangent cut’s $y$ axis then $x=0$.

$ \begin{aligned} & \therefore 0+\frac{y \sin \theta}{3}=1 \\ & \Rightarrow y=3 \operatorname{cosec} \theta \end{aligned} $

$\therefore$ Point of intersection at $y$ axis is $B(0,3 \operatorname{cosec} \theta)$.

$\therefore$ Length of $AB$

$ \begin{aligned} & =\sqrt{(4 \sec \theta-0)^2+(0-3 \operatorname{cosec} \theta)^2} \\ & =\sqrt{16 \sec ^2 \theta+9 \operatorname{cosec} e^2 \theta} \\ & =\sqrt{16\left(1+\tan ^2 \theta\right)+9\left(1+\cot ^2 \theta\right)} \\ & =\sqrt{25+16 \tan ^2 \theta+9 \cot ^2 \theta} \end{aligned} $

We know, $A M \geq G M$

$ \begin{aligned} & \therefore \frac{16 \tan ^2 \theta+9 \cot ^2 \theta}{2} \geq \sqrt{\left(16 \tan ^2 \theta\right)\left(9 \cot ^2 \theta\right)} \\ & \Rightarrow 16 \tan ^2 \theta+9 \cot ^2 \theta \geq 2(4 \tan \theta)(3 \cot \theta) \\ & \Rightarrow 16 \tan ^2 \theta+9 \cot ^2 \theta \geq 2 \times 4 \times 3 \\ & \Rightarrow 16 \tan ^2 \theta+9 \cot ^2 \theta \geq 24 \\ & \therefore A B=\sqrt{25+16 \tan ^2 \theta+9 \cot ^2 \theta} \\ & \geq \sqrt{25+24} \\ & \geq \sqrt{49} \\ & \geq 7 \end{aligned} $

$\therefore$ Minimum length of $A B=7$.