Differential Calculus
Differential Calculus
star_batch_jee_advanced_2025
Grade 12
Question:
Let $P(x_0, y_0)$ be a point on the curve $C : (x^2 - 11)(y + 1) + 4 = 0$ where $x_0, y_0 \in N$. If area of the triangle formed by the normal drawn to the curve 'C' at $P$ and the co-ordinate axes is $\left(\frac{a}{b}\right), a,b \in N$ then the least value of $(a - 6b)$.
Step-by-Step Solution
Key Concept: Find the point on the curve, compute the derivative to get the normal slope, then calculate the area of the triangle formed by the normal line and coordinate axes.
From $(x^2 - 11)(y + 1) = -4 - 2x^2$, rearrange to find the curve passes through $(3, 1)$. Computing the derivative: $y' = \frac{8x}{(x^2 - 11)^2}$, so $y'|_{x=3} = 6$ giving slope of normal $m_N = -\frac{1}{6}$. The equation of the normal line at $(3, 1)$ is $y - 1 = -\frac{1}{6}(x - 3)$, or $x + 6y = 9$. The area enclosed by the normal line with the axes is $\frac{1}{2} \times 9 \times \frac{3}{2} = \frac{27}{4}$.
Correct Answer: 3