# Math Help - Curves and tangents

1. ## Curves and tangents

I hope this is the right forum for this question.
Find the equation(s) of the tangent(s):
(i) from the point (1,0)
(ii) with gradient $-\frac{1}{2}$

(a) $y^2+4x=0$
ok for (i) for this equation, i tried differentiating the equation and plugging the values of x and y but i get -2/0 which definitely cannot be the case.
$2y\frac{dy}{dx}+4=0$
$\frac{dy}{dx}=-\frac{2}{y}$
the answer is supposed to be $y=\pm(x-1)$.
All I need would be some pointers, like where i should be looking. Thanks

2. Originally Posted by arze
I hope this is the right forum for this question.
Find the equation(s) of the tangent(s):
(i) from the point (1,0)
(ii) with gradient $-\frac{1}{2}$

(a) $y^2+4x=0$
ok for (i) for this equation, i tried differentiating the equation and plugging the values of x and y but i get -2/0 which definitely cannot be the case.
$2y\frac{dy}{dx}+4=0$
$\frac{dy}{dx}=-\frac{2}{y}$
the answer is supposed to be $y=\pm(x-1)$.
All I need would be some pointers, like where i should be looking. Thanks

After differentiating you have:

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

you seek a point where the derivative is $-1/2$, so we have at such a point:

y+4=0

or y=-4, and so x=-4. That is the point of tangency is (-4,-4)

However the line from (1,0) to (-4,-4) does not have gradient -1/2 so there can be no such line.

CB

CB

3. (i) and (ii) are separate, i need to find the equation of the tangent that passes through the point (1,0) this is the only condition.

4. Originally Posted by arze
I hope this is the right forum for this question.
Find the equation(s) of the tangent(s):
(i) from the point (1,0)
(ii) with gradient $-\frac{1}{2}$

(a) $y^2+4x=0$
ok for (i) for this equation, i tried differentiating the equation and plugging the values of x and y but i get -2/0 which definitely cannot be the case.
$2y\frac{dy}{dx}+4=0$
$\frac{dy}{dx}=-\frac{2}{y}$
the answer is supposed to be $y=\pm(x-1)$.
All I need would be some pointers, like where i should be looking. Thanks
(i) The point (1, 0) does not lie on the given curve. Do you mean that you need to find the equation of the tangent to $y^2 + 4x$ that passes through the point (1, 0)? In that case:

Let the point on the curve be $\left(- \frac{a^2}{4}, a \right)$. Then $m = - \frac{2}{a}$. But since the tangent passes through $\left(- \frac{a^2}{4}, a \right)$ and (1, 0) then $m = \frac{a}{-\frac{a^2}{4} - 1} = - \frac{4a}{a^2 + 4}$.

Therefore $- \frac{2}{a} = - \frac{4a}{a^2 + 4} \Rightarrow a = \pm 2$.

Case 1: a = 2

The point on the curve is (-1, 2) and m = -1. Therefore the equation of the tangent is $y - 2 = -1(x + 1) \Rightarrow y = -x + 1$.

Case 2: a = -2

The point on the curve is (-1, -2) and m = 1. Therefore the equation of the tangent is $y + 2 = 1(x + 1) \Rightarrow y = x - 1$.

5. Originally Posted by mr fantastic
(i) The point (1, 0) does not lie on the given curve. Do you mean that you need to find the equation of the tangent to $y^2 + 4x$ that passes through the point (1, 0)? In that case:

Let the point on the curve be $\left(- \frac{a^2}{4}, a \right)$. Then $m = - \frac{2}{a}$. But since the tangent passes through $\left(- \frac{a^2}{4}, a \right)$ and (1, 0) then $m = \frac{a}{-\frac{a^2}{4} - 1} = - \frac{4a}{a^2 + 4}$.

Therefore $- \frac{2}{a} = - \frac{4a}{a^2 + 4} \Rightarrow a = \pm 2$.

Case 1: a = 2

The point on the curve is (-1, 2) and m = -1. Therefore the equation of the tangent is $y - 2 = -1(x + 1) \Rightarrow y = -x + 1$.

Case 2: a = -2

The point on the curve is (-1, -2) and m = 1. Therefore the equation of the tangent is $y + 2 = 1(x + 1) \Rightarrow y = x - 1$.
Ahh.. two questions not two conditions, why did I not think of that...

CB

6. Originally Posted by CaptainBlack
Ahh.. two questions not two conditions, why did I not think of that...

CB
Well, it was a bit ambiguous. On another world line, I posted what you did and vice versa.

7. Originally Posted by mr fantastic
(i) The point (1, 0) does not lie on the given curve. Do you mean that you need to find the equation of the tangent to $y^2 + 4x$ that passes through the point (1, 0)? In that case:

Let the point on the curve be $\left(- \frac{a^2}{4}, a \right)$. Then $m = - \frac{2}{a}$. But since the tangent passes through $\left(- \frac{a^2}{4}, a \right)$ and (1, 0) then $m = \frac{a}{-\frac{a^2}{4} - 1} = - \frac{4a}{a^2 + 4}$.

Therefore $- \frac{2}{a} = - \frac{4a}{a^2 + 4} \Rightarrow a = \pm 2$.

Case 1: a = 2

The point on the curve is (-1, 2) and m = -1. Therefore the equation of the tangent is $y - 2 = -1(x + 1) \Rightarrow y = -x + 1$.

Case 2: a = -2

The point on the curve is (-1, -2) and m = 1. Therefore the equation of the tangent is $y + 2 = 1(x + 1) \Rightarrow y = x - 1$.
Thanks! I hadn't though of that.