Radius = perpendicular distance from (1, -3) to the given line 3x - 4y - 5 = 0.
i.e. $\left|\frac{3+12-5}{\sqrt{5^2}}\right|=2.$

Since, circles are passing through the point (–1, 1) and touching X-axis, the equationof circles can be written as - 
$(h+1{)}^2+(K–1{)}^2=K^2$
$h^2+2h+2–2K=0$
If h is real the $\Delta \ge 0$
4- 4.1. (2 – 2K)  $\ge 0$
= 4 - 8 + 8k $\ge 0$
= 8k $\ge 4$
= $k\ge \frac{1}{2}$

It is a fundamental concept. The end points of latus rectum of the parabola $x^2=4ay$ are (-2a , a) , (2a, a).

$x^2=-8y\Rightarrow a=-2.$ So , focus = (0,-2)

Ends of latus rectum = (4,-2) , (-4,-2) .

Trick: Since the ends of latus rectum lie on parabola , so only points (-4,-2) and (4,-2) satsify the parabola. 

$y_1=3x_1$ . According to given condition $9x_1^2=36x_1$ 

$\Rightarrow x_1=4,0\Rightarrow y_1=12,0$ 

Hence the points are (0,0) and (4,12). 

Given that a = 3
$LatusRectum=\frac{2b^2}{a}=\frac{16}{9}\frac{2}{3}{b}^2=\frac{16}{9}{b}^2=\frac{8}{3}$

The standard equation of the ellipse is 

$\frac{x^2}{9}+\frac{3y^2}{8}=1$

a = 4 , vertex = (0,0) , focus = (0,-4) . 

The given ellipse is
$\frac{x^2}{16}+\frac{y^2}{8}=1a^2=16;b^2=8\text{Focal length, c =}\sqrt{16-8}=\sqrt{8}$

The foci of the ellipse lie on the major axis. The greatest perpendicular distance from the major axis to any point on the ellipse is the length of the semi-minor axis.

For the triangle PST,
$ST=2S=2\sqrt{8}\text{Max Area =}\frac{1}{2}\times ST\times b=\frac{1}{2}\times 2\sqrt{8}\times \sqrt{8}=\text{8 sq. units}$

$12x^2+7y^2=84\frac{x^2}{7}+\frac{y^2}{12}=1a^2=7;b^2=12Focallength,c=\sqrt{b^2-a^2}c=\sqrt{12-7}=\sqrt{5}e=\frac{c}{a}=\sqrt{\frac{5}{12}}$

Given that the sum of the distances from (x,y) to the points (2,0) and (-2,0) is 8.
Let the  equation of the ellipse be $\frac{X^2}{a^2}+\frac{Y^2}{b^2}=1$
$\therefore 2a=8\Rightarrow a=4;c=2b^2=a^2-c^2\Rightarrow b^2=16-4=12$

The equation of  the ellipse is $\frac{X^2}{16}+\frac{Y^2}{12}=1$

If (x,3) lies on the ellipse, then, 
$\frac{x^2}{16}+\frac{3^2}{12}=1\Rightarrow \frac{x^2}{16}=\frac{1}{4}\Rightarrow x^2=4x=2\text{[only positive square root is considered]}$