Practicing Success
The normal to the curve $y(x-2)(x-3)=x+6$ at the point where the curve intersects the y-axis passes through the point |
$\left(-\frac{1}{2},-\frac{1}{2}\right)$ $\left(\frac{1}{2}, \frac{1}{2}\right)$ $\left(\frac{1}{2},-\frac{1}{3}\right)$ $\left(\frac{1}{2}, \frac{1}{3}\right)$ |
$\left(\frac{1}{2}, \frac{1}{2}\right)$ |
The equation of the curve is $y(x-2)(x-3)=x+6$ ......(i) It intersects y-axis at x = 0. Putting x = 0 in (i), we obtain y = 1. So, the point where (i) cuts y-axis has the coordinates (0, 1). From (i), we obtain $y=\frac{x+6}{x^2-5 x+6}$ Differentiating with respect to x, we obtain $\frac{d y}{d x}=\frac{\left(x^2-5 x+6\right)-(x+6)(2 x-5)}{\left(x^2-5 x+6\right)^2}$ ∴ $\left(\frac{d y}{d x}\right)_{(0,1)}=\frac{6-(-30)}{36}=1$ The equation of the normal at (0, 1) is $y-1=-1(x-0)$ or, $x+y-1=0$ Clearly, it passes through $\left(\frac{1}{2}, \frac{1}{2}\right)$. |