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C is a graph with equation [math]x^2-8y^2=k[/math], where k is a positive integer.
L is a straight line with equation [math]4y=x-1[/math].
L and C intersect at points A and B.
The x-coordinate of point A is -7.
(a) Work out the value of k.
(b) Work out the coordinates of point B.
Level
GCSE
Exam Board
Edexcel
- Najma asked 1 week ago
- last edited 1 week ago
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Best Answer
We are given:
\[ x^2-8y^2=k \]
and the straight line
\[ 4y=x-1 \]
The curves intersect at points A and B, and the x-coordinate of A is -7.
(a) Work out the value of k
Step 1: Find the coordinates of A
Substitute x=-7 into 4y=x-1:
\[ 4y=-7-1=-8 \]
\[ y=-2 \]
Therefore, A=(-7,-2)
Step 2: Substitute into the curve equation.
\[ x^2-8y^2=k \]
\[ (-7)^2-8(-2)^2=k \]
\[ 49-8(4)=k \]
\[ 49-32=k \]
\[ \boxed{k=17} \]
(b) Work out the coordinates of point B
Step 1: Substitute the line equation into the curve.
From 4y=x-1,
\[ y=\frac{x-1}{4} \]
Substitute into x^2-8y^2=17:
\[ x^2-8\left(\frac{x-1}{4}\right)^2=17 \]
Simplify:
\[ x^2-\frac{(x-1)^2}{2}=17 \]
Multiply everything by 2:
\[ 2x^2-(x-1)^2=34 \]
Expand:
\[ 2x^2-(x^2-2x+1)=34 \]
\[ x^2+2x-1=34 \]
\[ x^2+2x-35=0 \]
Step 2: Factorise.
\[ (x+7)(x-5)=0 \]
Therefore,
\[ x=-7 \quad \text{or} \quad x=5 \]
Since point A has x=-7, point B has
\[ x=5 \]
Step 3: Find the y-coordinate of B.
Using 4y=x-1:
\[ 4y=5-1=4 \]
\[ y=1 \]
Hence, the coordinates of point B are
\[ \boxed{B=(5,1)} \]
- Najma answered 1 week ago
- last edited 1 week ago
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