Practicing Success
The sum of the squares of the perpendicular drawn from the points $(0,1)$ and $(0,-1)$ to any tangent to a curve is 2 . The equation of the curve, is |
$2 y=C(x+2)$ $y=C(x \pm 1)$ $y=C(x+2)$ $y=C(x \pm 2)$ |
$y=C(x \pm 1)$ |
The equation of any tangent to a curve $y=f(x)$ is $Y-y=\frac{d y}{d x}(X-x)$ or, $X \frac{d y}{d x}-Y+\left(y-x \frac{d y}{d x}\right)=0$ ....(i) It is given that $\left|\frac{-1+y-x \frac{d y}{d x}}{\sqrt{\left(\frac{d y}{d x}\right)^2+1}}\right|^2+\left|\frac{1+y-x \frac{d y}{d x}}{\sqrt{\left(\frac{d y}{d x}\right)^2+1}}\right|^2=2$ $\Rightarrow 2\left\{\left(y-x \frac{d y}{d x}\right)^2+1\right\}=2\left\{1+\left(\frac{d y}{d x}\right)^2\right\}$ $\Rightarrow \left(y-x \frac{d y}{d x}\right)^2=\left(\frac{d y}{d x}\right)^2$ $\Rightarrow y-x \frac{d y}{d x}= \pm \frac{d y}{d x}$ $\Rightarrow \frac{1}{x \pm 1} d x=\frac{1}{y} d y$ $\Rightarrow \log (x \pm 1)=\log y+\log C$ $\Rightarrow \log (x \pm 1)=\log y+\log C$ $\Rightarrow C y=x \pm 1$ or, $y=k(x \pm 1)$ |