Q110 / 180 · COMEDK 2025 (25 May Shift) · Mathematics
A straight line is drawn through a given point $P(1,4)$. The least value of the sum of the intercepts on the co-ordinate axes is
- Correct answerA9
- B3
- C5
- D8
Solution
Let the equation of the line passing through $P(1, 4)$ be $\dfrac{x}{a} + \dfrac{y}{b} = 1$, where $a$ and $b$ are the x-intercept and y-intercept respectively.
Since the line passes through $(1, 4)$, we have $\dfrac{1}{a} + \dfrac{4}{b} = 1$.
We want to minimize the sum of intercepts $S = a + b$.
From the equation $\dfrac{1}{a} + \dfrac{4}{b} = 1$, we can express $b$ in terms of $a$: $\dfrac{4}{b} = 1 - \dfrac{1}{a} = \dfrac{a-1}{a} \Rightarrow b = \dfrac{4a}{a-1}$.
The sum $S = a + \dfrac{4a}{a-1}$. To minimize $S$, we differentiate with respect to $a$: $\dfrac{dS}{da} = 1 + \dfrac{(a-1)4 - 4a(1)}{(a-1)^2} = 1 - \dfrac{4}{(a-1)^2}$.
Setting $\dfrac{dS}{da} = 0$, we get $(a-1)^2 = 4$, so $a-1 = 2$ or $a-1 = -2$. Since intercepts must be positive for a minimum sum in the first quadrant, $a = 3$.
Substituting $a = 3$ into the expression for $b$: $b = \dfrac{4(3)}{3-1} = \dfrac{12}{2} = 6$.
The sum of intercepts is $S = a + b = 3 + 6 = 9$.
Answer: 9
Since the line passes through $(1, 4)$, we have $\dfrac{1}{a} + \dfrac{4}{b} = 1$.
We want to minimize the sum of intercepts $S = a + b$.
From the equation $\dfrac{1}{a} + \dfrac{4}{b} = 1$, we can express $b$ in terms of $a$: $\dfrac{4}{b} = 1 - \dfrac{1}{a} = \dfrac{a-1}{a} \Rightarrow b = \dfrac{4a}{a-1}$.
The sum $S = a + \dfrac{4a}{a-1}$. To minimize $S$, we differentiate with respect to $a$: $\dfrac{dS}{da} = 1 + \dfrac{(a-1)4 - 4a(1)}{(a-1)^2} = 1 - \dfrac{4}{(a-1)^2}$.
Setting $\dfrac{dS}{da} = 0$, we get $(a-1)^2 = 4$, so $a-1 = 2$ or $a-1 = -2$. Since intercepts must be positive for a minimum sum in the first quadrant, $a = 3$.
Substituting $a = 3$ into the expression for $b$: $b = \dfrac{4(3)}{3-1} = \dfrac{12}{2} = 6$.
The sum of intercepts is $S = a + b = 3 + 6 = 9$.
Answer: 9
Hence, the correct answer is option: 'A'.