Question

The equation of the tangent at $(x_1,y_1)$ to a curve $y^2=4ax$ at a point $(x_1,y_1)$ is given by $y{y}_1=2a(x+x_1)$

• A. True
• B. False

Answer 1

The correct option is A

True

The situation can be repersented as below

We know that equation of tangent at $(x_1,y_1)$ to a curve y=f(x) is given by

$y-y_1={\left[\frac{dy}{dx}\right]}_{(x_1,y_1)}(x-x_1)$

curve is

$y^2=4ax$

$2y.\frac{dy}{dx}=4a$

$\frac{dy}{dx}=\frac{2a}{y}$

$\Rightarrow {\left[\frac{dy}{dx}\right]}_{(x_1,y_1)}=\frac{2a}{y_1}$

So the equation of tangent to $y^2=4axat(x_1,y_1)$ is given by

$y-y_1=\frac{2a}{y_1}(x-x_1)$

$y{y}_1-y_1^2=2ax-2a{x}_1$

substracting $2a{x}_1$ from both sides

$y{y}_1-y_1^2-2a{x}_1=2ax-4a{x}_1$

$\Rightarrow y{y}_1-2a{x}_1=2ax(since{y}_1^2=4a{x}_1)$

$\Rightarrow y{y}_1=2a(x+x_1)$

$\therefore$ The given statement is true