F.5 Maths Module 2 question

x^2 + 4y^2 = 68
Differentiate both sides w.r.t. x, we get
2x + 8y dy/dx = 0
==> dy/dx = -x/(4y)
So, the slope of the normal at any point on this curve is 4y/x.
Suppose (x, y) is the required point on this curve, then
(y - 3)/(x - 0) = 4y/x
==> x(y - 3) = 4xy
==> x(y - 3) - 4xy = 0
==> x(y - 3 - 4y) = 0
==> -3x(y + 1) = 0
==> x = 0 or y = -1
When x = 0,
4y^2 = 68
==> y = √17 or y = -√17When y = -1,x^2 + 4(-1)^2 = 68==> x = 8 or -8equation of normal passes thru' (0, 3) and (0, √17) is x = 0;equation of normal passes thru' (0, 3) and (0, -√17) is x = 0;equation of normal passes thru' (0, 3) and (8, -1) is :(y - 3)/(x - 0) = 4*(-1)/8==> x + 2y - 6 = 0equation of normal passes thru' (0, 3) and (-8, -1) is :(y - 3)/(x - 0) = 4*(-1)/(-8)==> x - 2y + 6 = 0
Therefore, the equation of normals are :(i) x = 0(ii) x + 2y - 6 = 0(iii) x - 2y + 6 = 0
(Totally there are 3 normals pass thru' (0, 3))

x^2+〖4y〗^2=68

2x+4(2y)dy/dx=0

dy/dx=(-2x)/8y

dy/dx=(-x)/4y

When x=0, y=3

dy/dx=(-0)/3=0

Equation of the normal to C:

y-3=(-1)/0(x-0)

y=3

Refer to : http://www.mathportal.org/formulas/analytic-geometry/conic.php

to get the equation of tangent.
The ellipse is :
x²/(√68)² + y²/(√17)² = 1

Tangent at (0,3) is 0x/(√68)² + 3y/(√17)² = 1
3y = 17
slope of tangent = 0
slope of normal is -∞

equation of normal at (0,3) is x=0;