Applications of Derivatives
Tangent, Normal and Distance
nta_pyq_2023_apr
Grade 12
Question:
Let the tangent to the curve $x^2+2x-4y+9=0$ at the point $P(1,3)$ on it meet the $y$-axis at $A$. Let the line passing through $P$ and parallel to the line $x-3y=6$ meet the parabola $y^2=4x$ at $B$. If $B$ lies on the line $2x-3y=8$, then $(AB)^2$ is equal to _______.
Step-by-Step Solution
Key Concept: Find tangent at $P(1,3)$ using implicit differentiation, get point $A$ on $y$-axis. Then find the line through $P$ parallel to $x-3y=6$ and its intersection with $y^2=4x$.
Tangent at $P(1,3)$: $x+(x+1)-2(y+3)+9=0 \Rightarrow 2x-2y=-4$, giving $A=(0,2)$. Line through $P$ parallel to $x-3y=6$: $y-3=\frac{1}{3}(x-1)\Rightarrow 3y=x+8$. Intersection with $y^2=4x$: $B=(4,4)$ or $(16,8)$. Since $B$ lies on $2x-3y=8$, $B=(16,8)$. $(AB)^2=(0-16)^2+(2-8)^2=256+36=292$.
Correct Answer: 292