Question (1 Viewer)

[sonnylongbottom]

1. Find the two equations of the tangents of gradient 1/2 to ellipse x^2+6y^2=15

2. The point (x=3, y>0) lies on the ellipse x^2/25+y^2/9=1 Find equations of tangent and normal at P

Would love some help

 

[Affinity]

1.) differentiate implicitly to get dy/dx = - x/(6y) then solve dy/dx = 0 with x^2 + 6y^2 = 15 which should give you (x,y) = (3,-1) pr (-3,1) etc.

2.) Q2 is easier than Q1, find y then continue.

 

[vds700]

1. Find the two equations of the tangents of gradient 1/2 to ellipse x^2+6y^2=15

2. The point (x=3, y>0) lies on the ellipse x^2/25+y^2/9=1 Find equations of tangent and normal at P

Would love some help

1. You need to derive the general tangent formula for an ellipse. Let y = mx + k be a tangent to the ellipse at any point. Sub into the equation for the ellipse, expand and simplify, groupiong the x^2, x and constant terms so you have a quadratic equation in x. Now for a tangent, delta = 0. So find delta, make it = 0, then u should be able to solve for k. Then u can just sub x = 1/2 in

 

[sonnylongbottom]

1.) differentiate implicitly to get dy/dx = - x/(6y) then solve dy/dx = 0 with x^2 + 6y^2 = 15 which should give you (x,y) = (3,-1) pr (-3,1) etc.

2.) Q2 is easier than Q1, find y then continue.

Hey i dont get the ""solve dy/dx = 0 with x^2 + 6y^2 = 15"bit ? i implicitly differentiated and got -x/6y...

 

[3unitz]

Hey i dont get the ""solve dy/dx = 0 with x^2 + 6y^2 = 15"bit ? i implicitly differentiated and got -x/6y...

he meant dy/dx = 1/2,

-x/6y = 1/2 ----------(1)
x^2 + 6y^2 = 15 ----------(2)

solve (1) and (2) simultaneously.

this will get you the 2 points on the ellipse where the gradients of the tangents are 1/2.

 

[sonnylongbottom]

cheers

 

[3unitz]

1. You need to derive the general tangent formula for an ellipse. Let y = mx + k be a tangent to the ellipse at any point. Sub into the equation for the ellipse, expand and simplify, groupiong the x^2, x and constant terms so you have a quadratic equation in x. Now for a tangent, delta = 0. So find delta, make it = 0, then u should be able to solve for k. Then u can just sub x = 1/2 in

alternatively as vds700 suggested:

x^2 + 6y^2 = 15 ------(1)
y = x/2 + k ------(2)

sub (2) into (1):

x^2 + 6(x/2 + k)^2 = 15
x^2 + 6(x^2/4 + xk + k^2) = 15
x^2(1 + 6/4) + x(6k) + (6k^2 - 15) = 0

for tangent, delta = 0
(6k)^2 - 4(1 + 6/4)(6k^2 - 15) = 0
36k^2 - 10(6k^2 - 15) = 0
36k^2 - 60k^2 + 150 = 0
24k^2 = 150
k^2 = 25/4

.'. equation of tangents are:
y = x/2 + 5/2
y = x/2 - 5/2