# Thread: [SOLVED] How to solve this question?

1. ## [SOLVED] How to solve this question?

Find equation of normal at point (3,1) on curve $\displaystyle 2x^2 - xy + 3y^2 = 18$. The problem is with differentiating this $\displaystyle 2x^2 - xy + 3y^2 = 18$ part. How? Thanks

2. Originally Posted by mark1950
Find equation of normal at point (3,1) on curve $\displaystyle 2x^2 - xy + 3y^2 = 18$. The problem is with differentiating this $\displaystyle 2x^2 - xy + 3y^2 = 18$ part. How? Thanks

3. Thanks, Mr. Fantastic but I have a problem. Here is my take on the question :

I differentiate the whole thing to get,

$\displaystyle 4x - (x\frac{dy}{dx} + y) + 6y = 0$
$\displaystyle \frac{4x + 5y}{x} = \frac{dy}{dx}$

Then, I solved it to find the gradient :

$\displaystyle \frac{4(3) + 5(1)}{3} = \frac{dy}{dx}$

to get 17/3. Since they want me to find the normal, I solved the gradient to -3/17.

But when I try to find the equation, I got

17y = -3x + 26

which is way different then the answer, which is, 11y = 3x + 2. Am I wrong or is it the answer that's wrong? Thanks.

4. Originally Posted by mark1950
Thanks, Mr. Fantastic but I have a problem. Here is my take on the question :

I differentiate the whole thing to get,

$\displaystyle 4x - (x\frac{dy}{dx} + y) + 6y {\color{red} \frac{dy}{dx}} = 0$ Mr F says: Note the correction in red (so all that follows is wrong). This follows from the chain rule.

$\displaystyle \frac{4x + 5y}{x} = \frac{dy}{dx}$

Then, I solved it to find the gradient :

$\displaystyle \frac{4(3) + 5(1)}{3} = \frac{dy}{dx}$

to get 17/3. Since they want me to find the normal, I solved the gradient to -3/17.

But when I try to find the equation, I got

17y = -3x + 26

which is way different then the answer, which is, 11y = 3x + 2. Am I wrong or is it the answer that's wrong? Thanks.
Furthermore, since you only need the value of dy/dx and not the actual rule for dy/dx, my advice is to substitute (3, 1) into the line I corrected and then solve for dy/dx to get the required value.

5. Oh! How careless...lol. Thanks, Mr. F!