Question

Tangents are drawn to $x^2+y^2=16$ from the point $P(0,h)$. These tangents meet the x-axis at $A$ and $B$. If the area of $\Delta PAB$ is minimum, then $h=?$

• A. $12\sqrt{2}$
• B. $6\sqrt{2}$
• C. $8\sqrt{2}$
• D. $4\sqrt{2}$

Answer 1

The correct option is D $4\sqrt{2}$

Radius of circle $x^2+y^2=16$ is $4$ with centre at $(0,0)$

Let $M$ be $(x_1,y_1)$

$\therefore 2x+2y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=\frac{-x}{y}\therefore {\left|\frac{dy}{dx}\right|}_{atx=x_1,y_1}=\frac{-x_1}{y_1}$

Equation of line PB:

$(y-y_1)=\frac{-x_1}{y_1}(x-x_1)$

Put $(0,h)$$\Rightarrow h-y_1=\frac{-x_1}{y_1}(0-x_1)\Rightarrow h-y_1=\frac{{x_1}^2}{y_1}\Rightarrow h=\frac{{x_1}^2}{y_1}+y_1\Rightarrow h=\frac{{x_1}^2+{y_1}^2}{y_1}=\frac{16}{y_1}[\because x^2+y^2=16]$

Put $y=0$ in line PB:

$-y_1=\frac{-x_1}{y_1}(x-x_1)\Rightarrow \frac{{y_1}^2}{x_1}+x_1=xx=\frac{{x_1}^2+{y_1}^2}{x_1}=\frac{16}{x_1}$

$\therefore$ Coordinates of $B\equiv (\frac{16}{x_1},0)$

by Similarity,

Coordinates of $A\equiv (\frac{16}{x_1},0)$

$\therefore △PAB=\frac{1}{2}\times h\times \left(\frac{32}{x_1}\right)=\frac{16h}{x_1}=\frac{16}{x_1}\times \frac{16}{y_1}=\frac{{\left(16\right)}^2}{x_1{y}_1}=\frac{{\left(16\right)}^2}{x_1\sqrt{16-{x_1}^2}}$

For $△PAB$ to be minimum, $x_1\sqrt{16-{x_1}^2}$ has to be maximum.

$g\left(x\right)=x\sqrt{16-x^2}\Rightarrow g^{\prime }\left(x\right)=\sqrt{16-x^2}+\frac{x(-2x)}{2\sqrt{16-x^2}}=0\Rightarrow \sqrt{16-x^2}=\frac{x^2}{\sqrt{16-x^2}}\Rightarrow 16-x^2=x^2\Rightarrow x_1=2\sqrt{2}\Rightarrow y_1=\sqrt{16-x^2}=2\sqrt{2}\therefore h=\frac{16}{y_1}=\frac{16}{2\sqrt{2}}=4\sqrt{2}$

Hence, correct answer is $4\sqrt{2}$