Question

How many real tangents can be drawn from the point (4,3) to the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1.$ Find the equation of these tangents & angle between them.

Answer 1

Given point P=(4,3)

Ellipse

$S=\frac{x^2}{16}+\frac{y^2}{9}-1=0\because S,=\frac{16}{16}+\frac{9}{9}-1=>0\Rightarrow$

$\therefore (4,3)$Two tangents can be

drawn from the point P (4,3).

Equation of pair of tangents is $S{S}_1=T^2\Rightarrow (\frac{x^2}{16}+\frac{y^2}{9}-1)1={(\frac{4x}{16}+\frac{3y}{9}-)}^2$

$\Rightarrow \frac{x^2}{16}+\frac{y^2}{9}-1=\frac{x^2}{16}+\frac{y^2}{9}+1+\frac{xy}{6}-\frac{x}{2}-\frac{2y}{3}$

$\Rightarrow -xy+3x+4y-12=0\Rightarrow (4-x)(y-3)=0\Rightarrow x=4$ & y=3

and angle between them $\frac{\pi }{2}$