VCE Methods Question Thread!

[amanda4]

Find the equations of the lines that pass through the point (1, 7) and touch the parabola y = −3x^2 + 5x + 2
Hint: Form a quadratic equation and consider when the discriminant Δ is zero

[darkz]

Find the equations of the lines that pass through the point (1, 7) and touch the parabola y = −3x^2 + 5x + 2
Hint: Form a quadratic equation and consider when the discriminant Δ is zero

\[
\text{Let }m\text{ be the gradient of the equations}\\
\begin{aligned}\\
\implies y-7&=m(x-1)\\
&=mx-m+7...\boxed{1}\\
\text{ }\\
mx-m+7&=-3x^2+5x+2\\
0&=-3x^2+x(5-m)+m-5\\
\text{ }\\
\Delta &= (5-m)^2 - (-3)(4)(m-5)\\
&=m^2+2m-35\\
&=(m-5)(m+7)\\
m&=5,-7\\
\end{aligned}\\
\text{ }\\
\text{ Then substitute the values of m into }\boxed{1}\\
\text{ }\\
\begin{aligned}
y&=5x+2\\
y&=-7x+14
\end{aligned}
\]

[SmartWorker]

Hi,

Can you please help me with the following question: QU 3, 4, 7,

Thank You

[SmartWorker]

and these questions too please :

1(c,d,e), 2(b), 3(b,c,d), 4(a,ii, b,c)

[SmartWorker]

And this: 5-6

sorry its not allowing me to post everything together

[Sine]

Please refrain from double/triple posting in the future if that is possible.

As for your questions, what particular aspects are troubling you and what processes have you tried?

[SmartWorker]

Please refrain from double/triple posting in the future if that is possible.

As for your questions, what particular aspects are troubling you and what processes have you tried?

For qu 7 I tired graphing and finding the distance. But I got the answer wrong.
The rest of it I did the first parts of the question, but did not understand the rest

[S_R_K]

Hi,

Can you please help me with the following question: QU 3, 4, 7,

Thank You

A diameter of a circle passes through its centre. This gives you enough information to find the equations of the lines.

and these questions too please :

1(c,d,e), 2(b), 3(b,c,d), 4(a,ii, b,c)

It's useful to recall some circle geometry: a tangent to a circle is perpendicular to its radius. This makes easy work of a lot of the questions you've mentioned.

[Sine]

For qu 7 I tired graphing and finding the distance. But I got the answer wrong.
The rest of it I did the first parts of the question, but did not understand the rest

For question 7 - complete the square twice (for both x and y) and rearrange your equation into the form of a circle (x-h)^2 + (y-k)^2 = r^2
Then graph the circle.

[Ionic Doc]

Hello
Just got introduced to functions and I am struggling on this question badly! 
Can anyone help? Thnx

The function f has rule f(x) =  a/x + b      such that f(1) = 3/2  and f(2) = 9

1. Find the values of a and b
2. State the implied domain of F

Im predicting that you need to do some sort of simultaneous equation?
not really sure
thnx in advance 

[guac]

Hello
Just got introduced to functions and I am struggling on this question badly! 
Can anyone help? Thnx

The function f has rule f(x) =  a/x + b      such that f(1) = 3/2  and f(2) = 9

1. Find the values of a and b
2. State the implied domain of F

Im predicting that you need to do some sort of simultaneous equation?
not really sure
thnx in advance 

f(1) = a/1 + b = 3/2
f(2) = a/2 + b = 9

From there it's just a simultaneous equation to find a and b for part 1, as you said
For part 2, think about what values of x would 'break' the function (hint: what number can you never divide by?)

Hope that helps!

[peachxmh]

Hey guys, could someone explain how to solve this question? (in attached photo)
I'm getting a sense that you need to sub each of the values into the equation, but I'm not sure how this finds the point of intersection as there's only one equation, and how do you know if it's a point of intersection once you sub the point in?

It's basically the family of lines part that's confusing me since all the lines are condensed in one equation and I'm not sure how to use the equation to find the point of intersection. Thanks in advance

[S_R_K]

Hey guys, could someone explain how to solve this question? (in attached photo)
I'm getting a sense that you need to sub each of the values into the equation, but I'm not sure how this finds the point of intersection as there's only one equation, and how do you know if it's a point of intersection once you sub the point in?

It's basically the family of lines part that's confusing me since all the lines are condensed in one equation and I'm not sure how to use the equation to find the point of intersection. Thanks in advance

A better way to ask that question would be: "Which point is on a line of the form ... for all values of a, b in R \ {0}"?

[peachxmh]

A better way to ask that question would be: "Which point is on a line of the form ... for all values of a, b in R \ {0}"?

So in that case, I guess what I'm confused about is why a point is only on a line of that form for all a and b values if when subbed into the equation it equals zero? Because like generally if you are given the equation of just two lines (e.g. y=3x+2, y=x+3 - not a family of lines), a point of intersection is when you sub in an x value (for example) into both equations and you get the same y value answer from both equations. In this question though, they already give you the point of intersection, so you can't use this method and way of thinking?

Why does the point of intersection have to equal 0 when subbed into the equation - at least I'm assuming it does (because when you sub in each of the possible answers, all of them give an expression (e.g. 3b) whereas (D) just gives "true", because when (1,-4) is subbed in, it equals 0)

Sorry if what I'm saying is really confusing... I'm lowkey confused about what I'm confused about haha

[S_R_K]

So in that case, I guess what I'm confused about is why a point is only on a line of that form for all a and b values if when subbed into the equation it equals zero? Because like generally if you are given the equation of just two lines (e.g. y=3x+2, y=x+3 - not a family of lines), a point of intersection is when you sub in an x value (for example) into both equations and you get the same y value answer from both equations. In this question though, they already give you the point of intersection, so you can't use this method and way of thinking?

You can use the same way of thinking. A point (x, y) is on the line given by that equation if and only if the equation is true when the x and y values are subbed in.

Why does the point of intersection have to equal 0 when subbed into the equation - at least I'm assuming it does (because when you sub in each of the possible answers, all of them give an expression (e.g. 3b) whereas (D) just gives "true", because when (1,-4) is subbed in, it equals 0)

You've pretty much answered your own question here. (1, –4) is the point at which the family of lines intersect because it is the only point that makes the equation true, for any values of a and b. That's why your CAS returns "true" when you substitute in. In contrast, if substituting in a point gives you an expression like "3b", that means that point is on the line for some values of b but nor for others - hence, it is not a point through which every line in the family passes.