Question

Chap 5
7)a)

Answer 1

http://prntscr.com/4kted7

Answer 2

@ganeshie8

Answer 3

@ganeshie8

Answer 4

not sure of this
wil try after sometime

Answer 5

ok

Answer 6

@Miracrown

Answer 7

@kirbykirby

Answer 8

@skullpatrol

Answer 9

You should try equating both equations. This should give you any intersection points between the line and the ellipse. If here is only one solution, then you should establish that this is a tangent point, since a tangent line means that there is only one point of contact.

\(x+2y=4\implies x=4-2y\)

Now you can try substituting this \(x\) into the ellipse's equation:
\( (4-2y)^2+4y^2=8\\
16-16y+4y^2+4y^2=8\\
8y^2-16y+8=0\\
8y^2-8y-8y+8=0\\
8y(y-1)-8(y-1)=0\\
(y-1)(8y-8)=0\\
8(y-1)^2=0\\
\implies y=1\)

Now try substituting y=1 into the line/ellipse equation to get the value of x:

In the ellipse:
\( x^2+4(1)^2=8 \implies x^2=8-4=4 \\
x=\pm2\)

In the line equation:
\(x+2(1)=4 \implies x=2\)
We only get one x-solution when plugged into the line. If you try x = -2, you don't get y=1 from the line equation, so we disregard the solution x = -2.

Hence, there is only one point which is a valid solution, \(x=2, y=1\)

Answer 10

Thanks a lot @kirbykirby

Answer 11

@ganeshie8

Answer 12

nice :) thanks for tagging me, i was not getting notification when somebody replies :o

Answer 13

glad to help=]

Answer 14

@kirbykirby
I'm not sure what kind of ellise equation did you used..
Is it