Question 5
Given that $k$ is a non‑zero constant curve $C$ has equation $kx^2 – xy + (k + 1)x = 1$ straight line $l$ has equation $y = (k/2)x + 1$ The point $A$ is the only point that lies on both $C$ and $l.$
(a) Find the value of $k$ (6)
(b) Hence, find the coordinates of $A.$ (2)
Problem Solution
We are given:
- The curve $C$ with the equation: $$ kx^2 - xy + (k + 1)x = 1 $$
- The straight line $l$ with the equation: $$ y = \frac{k}{2}x + 1 $$
The point $A$ is the only point that lies on both $C$ and $l$.
Part (a): Find the value of $k$
Substitute $y = \frac{k}{2}x + 1$ from the equation of $l$ into $C$:
$$ kx^2 - x\left(\frac{k}{2}x + 1\right) + (k + 1)x = 1 $$Simplify:
$$ kx^2 - \frac{k}{2}x^2 - x + (k + 1)x = 1 $$Combine like terms:
$$ \frac{k}{2}x^2 + kx = 1 $$Factorize:
$$ x\left(\frac{k}{2}x + k\right) = 1 $$Divide through by $x$ (since $x \neq 0$):
$$ \frac{k}{2}x + k = \frac{1}{x} $$Multiply through by $x$:
$$ kx^2 + 2kx - 2 = 0 $$For $A$ to be the only point of intersection, the discriminant must be zero:
$$ \Delta = b^2 - 4ac $$Here, $a = k$, $b = 2k$, and $c = -2$:
$$ \Delta = (2k)^2 - 4(k)(-2) = 4k^2 + 8k $$Set $\Delta = 0$:
$$ 4k^2 + 8k = 0 $$Factorize:
$$ 4k(k + 2) = 0 $$Since $k \neq 0$, we have:
$$ k + 2 = 0 \quad \Rightarrow \quad k = -2 $$Part (b): Find the coordinates of $A$
Substitute $k = -2$ into the straight line equation:
$$ y = \frac{-2}{2}x + 1 \quad \Rightarrow \quad y = -x + 1 $$Substitute $k = -2$ into the curve equation:
$$ -2x^2 - xy - x = 1 $$Substitute $y = -x + 1$ into the curve equation:
$$ -2x^2 - x(-x + 1) - x = 1 $$Simplify:
$$ -2x^2 + x^2 - x - x = 1 $$ $$ - x^2 - 2x = 1 $$Rearrange:
$$ x^2 + 2x + 1 = 0 $$Factorize:
$$ (x + 1)^2 = 0 $$ $$ x = -1 $$Substitute $x = -1$ into $y = -x + 1$:
$$ y = -(-1) + 1 = 2 $$Thus, the coordinates of $A$ are:
$$ A = (-1, 2) $$
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