Name: Tangents to a Circle From an External Point
Uploaded: 2019-07-19
Duration: 3 min 5 s
Description: Do you know the equation of the tangents to a circle drawn from an external point? Let's find the equation involving the slope, which helps to find the equation of tangents.

Question

A parabola is drawn to pass through

A

and

B,

the ends of a diameter of a given circle of radius

a,

and to have as directrix a tangent to a concentric circle of radius

b;

then axes being

A B

and a perpendicular diameter, prove that the locus of the focus of the parabola is

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

.

Accepted Answer

Let focus of the parabola $S$ be $a,$

The circle $A B$ let the point of diameter are $M (q, 0), M (- q, 0)$ tangent $\equiv$ directrix of parabola

Also, we have a concentric circle $x^{2} + y^{2} = b^{2}$ whose tangent is directrix of the parabola

$∵$ We know the equation of tangent to the circle $y = m x + r \sqrt{m^{2} + 1} (r = b)$

So, $y = m x + b \sqrt{m^{2} + 1}$

or $m x - y + b \sqrt{m^{2} + 1} = 0 — (i) \to$ Directrix of parabola

We know in parabola $P S = P M$

$∵$ Parabola passes through point $P$ and also passing through the and point of circle $\frac{x^{2}}{b^{2}} + \frac{y^{2}}{b^{2} - a^{2}} = 1$ at $M$ . So, take $P (a, 0)$

Now, apply $P S = P M$

$\sqrt{{(h - a)}^{2} + k^{2}} = |\frac{m (a) - 1 (0) + b \sqrt{m^{2} + 1}}{\sqrt{m^{2} + 1}}| . . . (i i)$

Similarly,

Also $P S = P M^{1}$

So, $\sqrt{(h + a) + k^{2}} = |\frac{m (- a) - 1 \times (0) + b \sqrt{m^{2} + 1}}{\sqrt{m^{2} + 1}}| . . . (i i i)$

Adding equations $(i i)$ and $(i i i)$

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

$\Rightarrow \sqrt{{(h - a)}^{2} + k^{2}} + \sqrt{{(h + a)}^{2} + k^{2}} = \frac{a m + b \sqrt{m^{2} + 1} - a m + b \sqrt{m^{2} + 1}}{\sqrt{m^{2} + 1}}$

$\Rightarrow \sqrt{{(h - a)}^{2} + k^{2}} + \sqrt{{(h + a)}^{2} + k^{2}} = 2 b$

$\Rightarrow \sqrt{{(h - a)}^{2} + k^{2}} = 2 b - \sqrt{{(h + a)}^{2} + k^{2}}$

Squaring both sides $h^{2} + a^{2} + 2 a h + k^{2} = 4 b^{2} + (h^{2} + a^{2} - 2 a h) + k^{2} - 4 b \sqrt{{(h - a)}^{2} + k^{2}}$

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

Equation both sides again $b^{4} + a^{2} h^{2} - 2 b^{2} a h = b^{2} (h^{2} + a^{2} - 2 a h + k^{2})$

$b^{2} (b^{2} + \frac{a^{2} h^{2}}{b^{2}}) = b^{2} (h^{2} + a^{2} + k^{2})$

$b^{2} - a^{2} - h^{2} (1 - \frac{a^{2}}{b^{2}}) + k^{2}$

$(b^{2} - a^{2}) = h^{2} (\frac{b^{2} - a^{2}}{b^{2}}) + k^{2}$

$1 = h^{2} (\frac{b^{2} - a^{2}}{b^{2}}) \cdot \frac{1}{(b^{2} - a^{2})} + \frac{k^{2}}{(b^{2} - a^{2})}$

So, $\frac{h^{2}}{b^{2}} + \frac{k^{2}}{(b^{2} - a^{2})} = 1$

Put $h$ as $x$ and $k$ as $y$ .

So, $\frac{x^{2}}{b^{2}} + \frac{y^{2}}{b^{2} - a^{2}} = 1$ .

Important Questions on Circle

Learn and Prepare for any exam you want !

A parabola is drawn to pass through A and B, the ends of a diameter of a given circle of radius a, and to have as directrix a tangent to a concentric circle of radius b; then axes being AB and a perpendicular diameter, prove that the locus of the focus of the parabola is x2b2+y2b2-a2=1.

Important Questions on Circle

Learn and Prepare for any exam you want !