Write down the equation of the tangent to $x^2-3y^2=4y$ at the point $(x_1,y_1)$.

Question

Write down the equation of the tangent to $x^2-3y^2=4y$ at the point $(x_1,y_1)$.

The textbook gives the answer as $xx_1-3yy_1=2(y+y_1)$ and I'm not sure how it got there.

Okay so I differentiated $x^2-3y^2=4y$ to get:

$\frac{dy}{dx}=\frac{x}{3y+2}$

Then to find the equation of the tangent I usee $y-y_1=m(x-x_1)$

$y-y_1=\frac{x_1}{3y_1+2}(x-x_1)$

So I'm unsure how $y-y_1=\frac{x_1}{3y_1+2}(x-x_1)$ can be simplified to $xx_1-3yy_1=2(y+y_1)$ which is what the textbook says.

Help much appreciated.

Answer 1

We differentiate

$$ x^2 - 3y^2 = 4y$$

To get, $$ 2x -6y \frac{dy}{dx} = 4 \frac{dy}{dx} $$

To arrive at,

$$ 2x = \frac{dy}{dx}(4 + 6y) $$

Yielding

$$ \frac{x}{2 + 3y} = \frac{dy}{dx} $$

So at the target point as you rightly noted the slope will be

$$ \frac{x_1}{2 + 3y_1} $$

The equation of the point then will be

$$ \frac{x_1}{2 + 3y_1}(x - x_1) +y_1 = y $$

Which matches up with your solution. So now how to simplify this? We can multiply through by the denominator

$$ x_1(x - x_1) + y_1(2 + 3y_1) = y(2 + 3y_1) $$

$$ xx_1 - x_1^2 + 2y_1 + 3y_1^2 = 2y + 3yy_1 $$

Recall that $$ x^2 - 3y^2 = 4y$$ from the equation and thus the term

$$ -x_1^2 + 3y_1^2 $$ can be rightly rewritten as $$-4y_1$$

$$ xx_1 - 4y_1 + 2y_1 = 2y + 3yy_1 $$

$$ xx_1 - 3yy_1 = 2y_1 + y $$

$$ xx_1 - 3yy_1 = 2(y + y_1) $$

The trick, is knowing to resubtitute the definition of the equation you got from the start. It's a bit subtle.

Answer 2

Your equation has a fraction, the one you want to get to does not. Your equation has parentheses, the textbook answer does not.

Therefore, take your equation, multiply both sides by $3y_1+2$, expand the parentheses everywhere, and you should be a lot closer.

Answer 3

You got $y-y_1=\frac{x_1}{3y_1+2}(x-x_1)$ which is correct.

Now simplify $(y-y_1)(3y_1+2)=x_1(x-x_1)$ then $3yy_1-3y_1^2+2y-2y_1=xx_1-x_1^2$. By rearranging we get $xx_1-3yy_1-2y=x_1^2-3y_1^2-2y_1$. Adding $-2y_1$ both sides $\Rightarrow$ $xx_1-3yy_1-2(y+y_1)=x_1^2-3y_1^2-4y_1=0$ (Because $(x_1,y_1)$ is on the given curve $x^2-3y^2-4y=0$.) Hence the provided answer by your textbook.