Points of intersection of line $y=m x+c$ with the ellipse

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Putting $y=m x+c$ in the equation of ellipse gives

$\frac{x^2}{a^2}+\frac{(m x+c)^2}{b^2}=1$

$\Leftrightarrow b^2 x^2+a^2\left[m^2 x^2+2 m c x+c^2\right]=a^2 b^2$

$\left(a^2 m^2+b^2\right) x^2+2 a^2 m c x+a^2\left(c^2-b^2\right)=0$

This is a quadratic equation in $x$ with discriminant $D=\left(2 a^2 m c\right)^2-4\left(a^2 m^2+b^2\right) a^2\left(c^2-b^2)\right.$

that is $D=4 a^4 m^2 c^2-4 a^4 a^2 c^2+$

$4 a^4 m^2 b^2-4 a^2 b^2 c^2 +4 a^2 b^4$

$=4 a^2 b^2\left(a^2 m^2-c^2+b^2\right)$

Two points of intersection if $D>0$ that is

$c^2

One points of intersection if $D=0$ that is

$c^2=a^2 m^2+b^2$

No points of intersection if $D<0$ that is $c^2>a^2 m^2+b^2$

conclusion: The line $y=m x+c$ is tangent to the ellipse

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

if and only if $c^2=a^2 m^2+b^2$

Length of the chord joining $\left(x_1, y_1\right)$ and $\left(x_2, y\right.$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

The equation of the line joining $(x_1, y_1)$ & $(x_2, y_2)$ is

$y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$

Length of the chord

$l=\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}$

$=\sqrt{\left(x_1-x_2\right)^2+\left(m x_1+c-m x_2-c\right)^2}$

$=\sqrt{1+m^2}\left|x_1-x_2\right|$

$\left(x_1-x_2\right)^2=\left(x_1+x_2\right)^2-4 x_1 x_2$

$=\frac{4 a^4 m^2 c^2}{\left(a^2 m^2+b^2\right)^2}-4 \frac{a^2\left(c^2-b^2\right)}{a^2 m^2+b^2}$

$=\frac{4 a^2 b^2\left(a^2 m^2+b^2-c^2\right)}{\left(a^2 m^2+b^2\right)^2}$

$\therefore\left|x_1-x_2\right|=\frac{2 a b \sqrt{a^2 m^2+b^2-c^2}}{a^2 m^2+b^2}$

$l=\sqrt{1+m^2}\left|x_1-x_2\right|$

Equation of the tangent at the point $(x_1,y_1)$ to the ellopse $\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$ :

let the slope of the tangent at $(x_1, y_1)$be m. Then the equation of the tangent is $y-y, = m (x-x_1)$

that is $y= mx +(y_1-mx_1)$

put $c = y_1-mx_1$

We know that the line $y = mx+c$ is tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ if and only $c^2 = a^2m^2+b^2$

$\Rightarrow (y_1 - mx_1)^2 = a^2m^2+b^2$

$\Rightarrow (x_1^2 - a^2) m^2 - 2x_1 y_1 m + y^2 - b^2 = 0$

D = $4x_1^2 y_1^2 - 4 (x_1^2 - a^2) (y_1^2 - b^2)$

that is $D =4\left(a^2 y_1^2+b^2 x_1^2-a^2 b^2\right)$

$=4 a^2 b^2\left(\frac{x_1^2}{a^2}+\frac{y_1^2}{b}-1\right)=0$

$\therefore \quad m =\frac{2 x_1 y_1}{2\left(x_1^2-a^2\right)}=\frac{x_1 y_1}{x_1^2-a^2} .$

$(\therefore (x_1, y_1)$ lies on the ellipse)

So, the equation of the tangent line is

$\Leftrightarrow y-y_1=\frac{x_1 y_1}{x_1^2-a^2}(x-x_1)$

$\Leftrightarrow y-y_1=\frac{x_1 y_1}{x_1^2-a^2} x-\frac{x_1^2 y_1}{x_1^2-a^2}$

$\Leftrightarrow y y_1-y_1^2=\frac{x_1 y_1^2}{x_1^2-a^2} x-\frac{x_1^2 y_1^2}{x_1^2-a^2}$

$\Leftrightarrow y y_1=\frac{y_1^2}{x_1^2-a^2}\left(x x_1\right)+y_1^2 \left[1-\frac{x_1^2}{x_1^2-a^2}\right]$

$\Leftrightarrow y y_1 =\frac{y_1^2}{x_1^2-a^2}\left(x x_1\right)+y_1^2\left(\frac{-a^2}{x_1^2-a^2}\right)$

$=\frac{y_1^2}{x_1^2-a^2}\left[x x_1-a^2\right]$

Now,

$\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}=1 \Rightarrow b^2 x_1^2+a^2 y_1^2=a^2 b^2$

$\Rightarrow b^2\left(x_1^2-a^2\right)=-a^2 y_1^2$

$\Rightarrow \frac{y_1^2}{x_1^2-a^2}=-\frac{b^2}{a^2}$

$\therefore y y_1=\frac{-b^2}{a^2}\left[-x x_1-a^2\right]=-\frac{b^2}{a^2} x x_1+b^2$

$\Rightarrow \frac{x x_1}{a^2}+\frac{y y_1}{b^2}=1$

$\leftarrow$ Equation of tangent at $(x_1, y_1)$

Another way using Calculas

$\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$

Differentating w.r.t. $x_1$ we get

$\frac{2 x}{a^2}+\frac{2 y \frac{d y}{d x}}{b^2}=0 \Rightarrow \frac{d y}{d x}=-\frac{b^2}{a^2} \frac{x}{y}$

Recall : slope of tangent to a curve $y=f(x)$ at point $(x_1 , y_1)$ is $m=\left.\frac{d y}{d x}\right|_{(x, y)}$

$\therefore m=\frac{-b^2}{a^2} \frac{x_1}{y_1}$.

$\therefore$ Equation of the tangent is $y-y_1=\frac{-b^2}{a^2} \frac{x_1}{y_1}\left(x-x_1\right)$

that is $y-y_1=-\frac{b^2}{a^2} \frac{x_1}{y_1} x+\frac{b^2}{a^2} \frac{x_1^2}{y_1}$

that is $y y_1-y_1^2=\frac{-b^2}{a^2} x x_1+\frac{b^2}{a^2} x_1^2$

that is $y y_1+\frac{b^2}{a^2} x x_1=y_1^2+\frac{b^2}{a^2} x_1^2$

that is $\frac{y y_1}{b^2}+\frac{x x_1}{a^2}=\frac{y_1^2}{b^2}+\frac{x_1^2}{a^2}=1$

that is $\frac{x x_1}{a^2}+\frac{y y_1}{b^2}=1$

Equation of the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at the point $(x_1, y_1)$

We saw that the slope of tangent at $(x_1, y_1)$ is $\frac{-b^2}{a^2}\frac{x_1}{y_1}$

$\therefore$ Slope of the normal at $(x_1, y_1)$ is

$m = \frac{a^2y_1}{b^2x_1}$

$\therefore$ The equation is $y-y_1 = \frac{a^2y_1}{b^2x_1} (x-x_1)$

that is $\frac{x-x_1}{x_1/x^2} = \frac{y-y_1}{y_1/b^2}$

Parametric from of a general point on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} =1$ :

Any point $(x_1, y)$ on the ellipse satisfies $\left(\frac{x}{a}\right)^2 + \left(\frac{y}{b}\right)^2 =1$

Putting $\frac{x}{a} = \cos\theta$ and $\frac{y}{b} = \sin\theta$, we see that

$x = a \cos\theta$, $y = b \sin\theta$ gives any point on the ellipse for $0 \leq \theta<2 \pi$

$p =(a \cos\theta, b \sin\theta)$

We will call the angle $\theta$ to be the eccentric angle of the point $p$

The point $Q$ lies on the circle $x^2 + y^2 = a^2$ an has the same $x$ coordinate as the point $P$. So, any point on the ellipes $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ given by $(a \cos\theta, b \sin\theta)$ is on the vertical line thought the point on the circle $x^2+y^2 = a^2$ at an angle $\theta$ with the positive $x-$ axis

The ratio of the $y-$ coordinates of $P$ and $Q$ is $\frac{b \sin\theta}{a \sin\theta} = \frac{b}{a}$

$\frac{PM}{QM} = \frac{b}{a}$

that is PM = $\frac{b}{a} QM$

Determine the locus of the point of intersaction of tangents to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ which must at right angles

Solution: Equation of the tangent whose slope is $m$ is given by $y = mx + c$ with $c^2 = a^2 m^2 + b^2$

that is $y = mx + \sqrt{a^2 m^2 + b^2}$ - (1)

The equation of the tangent which is perpendicular to the above tangent is

$y = \frac{-1}{m} x +\sqrt{a^2(\frac{-1}{m})^2 + b^2}$

$y = \frac{-1}{m} x + \sqrt{ \frac{a^2}{m^2} + b^2}$ -(2)

If $(h_1 k)$ is the point of intersection of (1) & (2)

$k - mh=\sqrt{a^2+b^2m^2}$ -(3)

and $mk + h = \sqrt{a^2+b^2m^2}$ -(4)

We need to elliminate $m$ from (3) & (4) Squaring & adding (3) & (4) gives

$(k - mh)^2 + (mk + h)^2 = a^2 m^2 + b^2 +a^2 + b^2 m^2$

that is $(k^2 + h^2) (1 + m^2) = (a^2 + b^2) (1 + m^2)$

that is $h^2 + k^2 = a^2 + b^2$

$\therefore$ The locus is $x^2 + y^2 = a^2 + b^2$ $\leftarrow$ Circle conected at $(0, 0)$ & radius $\sqrt{a^2 + b^2}$

Thank you