Note: This is part 1st of a 2 part question, split as 2 separate questions here.

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

- If $y=f(x)$ then $\large\frac{dy}{dx}$$=f'(x)$ is the rate of change of $y$ w.r.t $x$
- $\large\frac{dy}{dx_{(x_1,y_1)}}$ is the slope of the tangent to the curve at the point $(x_1,y_1) $ on the curve. It is the slope of the curve at that point.
- The normal at a point $(x_1,y_1)$ on $y=f(x)$ is perpendicular to the tangent at $(x_1,y_1)$

$x^2+y^2-2x-4y+1=0$

Step 1:

Let $(x_1,y_1)$ be the point on the curve at which the tangent is $||$ to $x-axes =>m=0$

Therefore $\large\frac{dy}{dx_(x_1,y_1)}$$=0$

Step 2:

Differentiating $x^2+y^2-2x-4y+1=0$ with respect to x

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

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

$\large\frac{dy}{dx}=\frac{1-x}{y_1-2}$

$m=\large\frac{dy}{dx_{(x_1y_1)}}=\frac{1-x_1}{y_1-2}$

Step 3:

$1+y_1^2-2-4y_1+1=0$

$y_1^2-4y_1=0=>y_1(y_1-4)=0=>y_1=0,4$

The points are $(1,0) (1,4)$

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...