Application of Derivatives — Class 12 Maths Solution

Given, equation of curve which is

${x^2} + {y^2} - 2x - 4y + 1 = 0$ …….(i)

$\Rightarrow$ $2x + 2y\frac{{dy}}{{dx}} - 2 - 4\frac{{dy}}{{dx}} = 0$

$\Rightarrow$ $\frac{{dy}}{{dx}}(2y - 4) = 2 - 2x$

$\Rightarrow$ $\frac{{dy}}{{dx}} = \frac{{2(1 - x)}}{{2(y - 2)}}$

Since, the tangents are parallel to the Y-axis i.e., $\tan \theta = \tan {90^\circ } = \frac{{dy}}{{dx}}$.

Therefore,$\frac{{1 - x}}{{y - 2}} = \frac{1}{0}$
$\Rightarrow$ $y - 2 = 0$

$\Rightarrow$ $y = 2$

For $y = 2$ from Eq. (i), we get
${x^2} + {2^2} - 2x - 4 \times 2 + 1 = 0$

$\Rightarrow$ ${x^2} - 2x - 3 = 0$

$\Rightarrow$ ${x^2} - 3x + x - 3 = 0$

$\Rightarrow$ $x(x - 3) + 1(x - 3) = 0$

$\Rightarrow$ $(x + 1)(x - 3) = 0$

Therefore,$x = - 1,x = 3$

Therefore the required points are (-1,2) and (3,2) .