Statement – 1:Tangents PA and PB are drawn to parabola, where A and B are points of contacts of tangents. If mid-point of chord AB is M then line PM is perpendicular to tangent at vertex.
Statement 2:  Maximum distance of a normal drawn to ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b>0)$ at any point from its centre is a + b.
 Statement 3:  The normal to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at a point ‘θ’ is $\frac{ax}{\cos \theta }-\frac{by}{\sin \theta }=a^2-b^2$

Statement 4: The radius of the largest circle having center at focus of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b>0)$ and inscribed in it is a(1 – e) where e is eccentricity of ellipse.
Which of the following must be truth value of above statements in that order.

For hte statement 1

The tangents at the points $A\left(t_1\right),B\left(t_2\right)$

The point of intersection of tangents is $P\left(a{t}_1{t}_2,a\left(t_1+t_2\right)\right)$ and midpoint of $AB$ is $M\left(\frac{a\left({t_1}^2+{t_2}^2\right)}{2},a\left(t_2+t_2\right)\right)$
The tangent to the parabola at the vertex is $y-$ axis, and the line $PM$ is parallel to $x-$ axis,

Hence, $PM$ is perpendicular to the tangent at the vertex. 

-For the statement 2

The equation of normal to the ellipse is $\frac{ax}{\cos \vartheta }-\frac{by}{\sin \vartheta }=a^2-b^2$

The perpendicular distance from origin to the above line is $\frac{a^2-b^2}{\sqrt{a^{2}se{c}^2\vartheta +b^2\cos e{c}^2\vartheta }}$

To get the maximum value of the above value, let us consider the minimum value of denomiantor, it is $a+b$

Hence, the required value is $a-b$

For the statement 3, It is formula

For the statement 4

The equatio of the circle is ${\left(x-ae\right)}^2+y^2=r^2$ and this touches the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$-

Substitute $y^2=b^2\left(1-\frac{x^2}{a^2}\right)$

$x^2+a^2{e}^2-2aex+b^2-\frac{b^2}{a^2}{x}^2-r^2=0$

It is a quadratic equation, it has only one root

$\begin{array}{rcl}4a^2{e}^2& =& 4\left(1-\frac{b^2}{a^2}\right)\left(a^2{e}^2-r^2\right)\\ & =& 4e^2\left(a^2{e}^2-r^2\right)\\ a^2& =& a^2{e}^2-r^2\\ r& =& a\sqrt{e^2-1}\end{array}$

Hence, statement 1, Statement 2, Statement 3 are correct, but statement 4 is wrong.