Differentiation question pls help

[From: ] [author: ] [Date: 11-12-24] [Hit: ]

-x^2+2y^2 =152............

Set this equal to -3, we find that x = 3 and y = 1/2, or any x and y that maintain a ratio of 6:1.

Using the two values x = 6 and y = 1, we can find the normal line by using slope 1/3, or the negative multiplicative inverse of -3.

(y - 1) = 1/3(x - 6)
y = (1/3)x + 1

If the values were x = 12 and y = 2:
y = (1/3)x + 3

And so we can write: y = (1/3)x + c, where c is of the form 2k + 1, k being a non negative integer.

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x^2+2y^2 =152 .................(i)
2x +4y(dy/dx) = 0
dy/dx = -(x/2y)
Therefore slope of normal = (2y/x)...........(ii)
slope of the line 2x -6y -7 = is =1/3 .........(iii)
from (ii) and (iii)
2y/x = 1/3
6y = x.............................(iv)
To find point on curve solve (i) and (iv)
36y^2 +2y^2 = 152
38 y^2 = 152
y^2 = 4 or y = +/- 2
Therefore points are (2,1/3) and (-2,-1/3)
Hence equation of normal at (2,1/3) is
y-1/3 = 1/3(x-2)
3y -1 =x-2
x-3y =1 ...................Ans
Hence equation of normal at (-2,-1/3) is
y+1/3 = 1/3(x+2)
3y +1 =x+2
x-3y = -1 ...................Ans

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