Question

Tangents are drawn to the circle $x^2+y^2=50$ from a point ${}^{\prime }{P}^{\prime }$ lying on the $x-$axis. These tangents meet the $y-$axis at points '$P_1$' and '$P_2$'. Possible coordinates of ${}^{\prime }{p}^{\prime }$ so that area of triangle $P{P}_1{P}_2$ is minimum, is

• A. $(10,0)$
• B. $(10\sqrt{2},0)$
• C. $(-10\sqrt{2},0)$
• D. $(10\sqrt{3},0)$

Answer 1

Answer: (A), (C)

The correct options are
A $(10,0)$
C $(-10\sqrt{2},0)$

We have:

$S:x^2+y^2-50=0$

Let co-ordinates of P be$(x_1,y_1)=(p,0)$

Equation of pair of tangents from $(x_1,y_1)$is:

$S{S}_1=T^2$

$\Rightarrow (x^2+y^2-50)(x_1^2+y_1^2-50)$

$=(x{x}_1+y{y}_1-50{)}^2$

$\Rightarrow (x^2+y^2-50)(p^2+0-50)=(px-50{)}^2$

$\Rightarrow (x^2+y^2-50)(p^2-50)=(px-50{)}^2$

This pair of tangents meet y - axis at points

where x = 0

$\Rightarrow (y^2-50)(p^2-50)=(0-50{)}^2$

$\Rightarrow (y^2-50)=\frac{2500}{p^2-50}$

$\Rightarrow y^2=\frac{2500}{p^2-50}+50=\frac{2500+50p^2-2500}{p^2-50}$

$\Rightarrow y^2=\frac{50p^2}{p^2-50}$

$\Rightarrow y\pm \sqrt{\frac{50p^2}{p^2-50}}$

$=\pm \sqrt{\frac{50}{p^2-50}}|p|$

Hence We have $Q(0,\sqrt{\frac{50}{p^2-50}}|p|)$ and

$R(0,-\sqrt{\frac{50}{p^2-50}}|p|)$

Note that : Q and R are equidistant from

origin.

Using distance formula you will get QR as

$QR=2|p|\sqrt{\frac{50}{p^2-50}}$ which will be base of

triangle.

Height of triangle will be OP = |P|

$\Rightarrow$ Area $=A=\frac{1}{2}\times Base\times Height$

$\Rightarrow A=\frac{1}{2}\times 2|p|\sqrt{\frac{50}{p^2-50}}\times |p|$

$\Rightarrow A=|p{|}^2\sqrt{\frac{50}{p^2-50}}$

$\Rightarrow A^2=\frac{50p^4}{p^2-50}$

Differentiate with respect to p, we get:

$2A.\frac{dA}{dp}=\frac{200p^2(p^2-50)-50p^4\left(2p\right)}{(p^2-50{)}^2}$

For maxims/ minima, $\frac{dA}{dp}=0$

$\frac{200p^3(p^2-50)-50p^4\left(3p\right)}{(p^2-50{)}^2}=0$

$\Rightarrow 200p^3(p^2-50)-50p^4\left(2p\right)=0$

$\Rightarrow 50p^3\left[4\right(p^2-50)-2p^2]=0$

$\Rightarrow 4(p^2-50)-2p^2=0$

$\Rightarrow 4p^2-200-2p^2=0$

$\Rightarrow 2p^2=200$

$\Rightarrow p^2=100$

$\Rightarrow p=\pm 10$

[I leave checking double derivative for you.]

Hence coordinates of p are (10,0) or

(-10, 0) $\to$ Option (A,C)