Question - How to get this...Equation to normal??? (1 Viewer)

[Smile12345]

Hello All...

Could someone please help me with the following question??



Thanks in advance.

 

[unforlornedhope]

Can you please post the pic again? The picture can't be seen


Sent from my iPhone using Tapatalk

 

[Smile12345]

Sure ... I might have clicked 'edit' for a moment.

 

[Troller]

You can find the gradient of the tangent by using the first derivative, then you can find the gradient of the normal. After this find the normal equation at P, then solve simultaneously with the curve y=1/8x^2 this should leave you with 2 x values, 1 of them will be 2, the other will be the x coordinate of Q, sub this into the equation of the curve, and you will have the coordinates of Q

 

[Smile12345]

You can find the gradient of the tangent by using the first derivative, then you can find the gradient of the normal. After this find the normal equation at P, then solve simultaneously with the curve y=1/8x^2 this should leave you with 2 x values, 1 of them will be 2, the other will be the x coordinate of Q, sub this into the equation of the curve, and you will have the coordinates of Q

Thanks for your help... I'm not quite getting the right answer... Could you please explain further??

 

[Carrotsticks]

1. Find the equation of the normal.

2. Solve simultaneously with the quadratic.

3. The solutions of that quadratic are the x coordinates of where the normal intersects the parabola.

4. However, we know that one of these x coordinates is at P (which is x=2)

5. Sum of roots = A + 2

6. Solve for A.

 

[Smile12345]

1. Find the equation of the normal.

2. Solve simultaneously with the quadratic.

3. The solutions of that quadratic are the x coordinates of where the normal intersects the parabola.

4. However, we know that one of these x coordinates is at P (which is x=2)

5. Sum of roots = A + 2

6. Solve for A.

1. So equation to normal is: 2x + 2y - 5 = 0 which is arranged to y = -x + 5/2

This correct so far??

 

[Smile12345]

Thanks for your help Carrot.

 

[Carrotsticks]

y' = x/4 so when x=2, the gradient of the tangent is 1/2. This means the gradient of the normal is -2 (negative reciprocal).

y-1/2 = -2(x-2)

2y-1 = -4(x-2) (multiplied both sides by 2 to get rid of the fraction)

2y-1 = -4x + 8

4x + 2y - 9 = 0

 

[Smile12345]

y' = x/4 so when x=2, the gradient of the tangent is 1/2. This means the gradient of the normal is -2 (negative reciprocal).

y-1/2 = -2(x-2)

2y-1 = -4(x-2) (multiplied both sides by 2 to get rid of the fraction)

2y-1 = -4x + 8

4x + 2y - 9 = 0

Yup, whoops... I forgot to multiply the RHS!