Show that the equation of the tangent at the point $(x_1,y_1)$ to the curve with equation $x^2-2y^2-6y = 0$ is $xx_1-2yy_1-3(y+y_1)=0$
Alright I'm having a problem following an example from my textbook. I'll just show what I do follow and then where I get confused.
Okay so I know how to find the equation of the tangent from first getting the gradient by differentiation.
$x^2-2y^2-6y=0$
$\Rightarrow$ $2x-2(2y\frac{dy}{dx})-6\frac{dy}{dx}=0$
$\Rightarrow$ $\frac{dy}{dx} = \frac{x}{3+2y}$
The gradient of the tangent at the point $(x_1,y_1)$ is $\frac{x_1}{3+2y_1}$
The equation of the tangent is given by $y-y_1=\frac{x_1}{3+2y_1}(x-x_1)$ which can be written as:
$(y-y_1)(3+2y_1)=x_1(x-x_1)$
The textbook says that this simplifies down to
$xx_1-2yy_1-3(y+y_1)=x_1^2-2y_1^2-6y_1$
I know I'm doing something wrong cause when I expand $(y-y_1)(3+2y_1)=x_1(x-x_1)$ I don't get that.
I get:
$3y-3y_1+2yy_1-2y_1^2=xx_1-x_1^2$ and then rearranging that I get
$xx_1-2yy_1-3(y-y_1)=x_1^2-2y_1^2$
And that's completely different to my textbook's answer.
So if anyone can shed some light on how
$(y-y_1)(3+2y_1)=x_1(x-x_1)$ simplifies to $xx_1-2yy_1-3(y+y_1)=x_1^2-2y_1^2-6y_1$
then I would be extremely grateful.
P.S Sorry to ask such a basic question I'm quite new to learning this stuff
If you add $6y_1$ to both sides of the text's "simplified" equation, you get your equation. Yours is actually simpler. Your algebra is correct.