Question

A line $L: y = mx + 9$ meets $y$-axis at point $A$ and intersect $y^2 = 12x$ at the point $B(x_0, y_0)$, where $0 < y_0 < 15$.

The tangent to the parabola at point $B$ intersects the $y$-axis at $C(0, y_1)$. The slope $m$ of the line $L$ is chosen such that the area of the triangle $ABC$ is maximum.

Match each entry in List-I with correct entries in List-II.

	List-I		List-II
(P)	$y_0 + x_0$ is equal to	(1)	24
(Q)	$m$ is equal to	(2)	$\frac{1}{4}$
(R)	Slope of tangent is equal to	(3)	$\frac{1}{2}$
(S)	Maximum area of $\triangle ABC$ (in sq. units) is	(4)	18
		(5)	6

Answer 1

P-1, Q-2, R-3, S-4

Answer 2

The line $L: y = mx + 9$ meets the $y$-axis at point $A$. Setting $x=0$, we get $y = 9$. So, $A = (0, 9)$.

The line $L$ intersects the parabola $y^2 = 12x$ at point $B(x_0, y_0)$.

Since $B$ is on the line, $y_0 = mx_0 + 9$.

Since $B$ is on the parabola, $y_0^2 = 12x_0$.

From the parabola equation, $x_0 = \frac{y_0^2}{12}$. Substitute this into the line equation:

$y_0 = m \left(\frac{y_0^2}{12}\right) + 9$

$12y_0 = my_0^2 + 108$

$my_0^2 - 12y_0 + 108 = 0$.

The tangent to the parabola $y^2 = 12x$ at $B(x_0, y_0)$ is given by $yy_0 = 6(x+x_0)$.

This tangent intersects the $y$-axis at $C(0, y_1)$. Setting $x=0$, we get $y_1 y_0 = 6x_0$.

$y_1 = \frac{6x_0}{y_0}$. Substitute $x_0 = \frac{y_0^2}{12}$:

$y_1 = \frac{6(y_0^2/12)}{y_0} = \frac{y_0^2/2}{y_0} = \frac{y_0}{2}$.

So, $C = (0, y_0/2)$.

The vertices of $\triangle ABC$ are $A(0, 9)$, $B(x_0, y_0)$, and $C(0, y_0/2)$.

The base $AC$ lies on the $y$-axis. The length of the base $AC$ is $|9 - y_0/2|$.

The height of the triangle with respect to base $AC$ is the absolute value of the $x$-coordinate of $B$, which is $|x_0|$. Since $y_0 \in (0, 15)$, $y_0^2 = 12x_0 > 0$, so $x_0 > 0$. The height is $x_0$.

Area of $\triangle ABC = \frac{1}{2} \times AC \times x_0 = \frac{1}{2} |9 - y_0/2| x_0$.

Given $0 < y_0 < 15$, $y_0/2 < 15/2 = 7.5$. So $9 - y_0/2 > 9 - 7.5 = 1.5 > 0$.

Area $S = \frac{1}{2} (9 - y_0/2) x_0$.

Substitute $x_0 = \frac{y_0^2}{12}$:

$S(y_0) = \frac{1}{2} (9 - y_0/2) \frac{y_0^2}{12} = \frac{y_0^2}{24} (9 - y_0/2) = \frac{9y_0^2}{24} - \frac{y_0^3}{48} = \frac{3y_0^2}{8} - \frac{y_0^3}{48}$.

To maximize the area, we find the derivative with respect to $y_0$ and set it to zero:

$S'(y_0) = \frac{d}{dy_0} \left(\frac{3y_0^2}{8} - \frac{y_0^3}{48}\right) = \frac{6y_0}{8} - \frac{3y_0^2}{48} = \frac{3y_0}{4} - \frac{y_0^2}{16}$.

$S'(y_0) = 0 \Rightarrow \frac{3y_0}{4} - \frac{y_0^2}{16} = 0 \Rightarrow y_0 \left(\frac{3}{4} - \frac{y_0}{16}\right) = 0$.

Since $y_0 > 0$, we must have $\frac{3}{4} - \frac{y_0}{16} = 0 \Rightarrow \frac{y_0}{16} = \frac{3}{4} \Rightarrow y_0 = \frac{3}{4} \times 16 = 12$.

This value $y_0 = 12$ is in the range $(0, 15)$.

The second derivative is $S''(y_0) = \frac{3}{4} - \frac{2y_0}{16} = \frac{3}{4} - \frac{y_0}{8}$. At $y_0=12$, $S''(12) = \frac{3}{4} - \frac{12}{8} = \frac{3}{4} - \frac{3}{2} = -\frac{3}{4} < 0$, confirming a maximum.

The maximum area occurs when $y_0 = 12$.

Now we calculate the values for List-I using $y_0 = 12$.

$x_0 = \frac{y_0^2}{12} = \frac{12^2}{12} = 12$. So $B = (12, 12)$.

(P) $y_0 + x_0 = 12 + 12 = 24$. This matches List-II (1).

(Q) The point $B(12, 12)$ lies on the line $y = mx + 9$.

$12 = m(12) + 9 \Rightarrow 3 = 12m \Rightarrow m = \frac{3}{12} = \frac{1}{4}$. This matches List-II (2).

(R) The slope of the tangent to $y^2 = 12x$ at $(x_0, y_0)$ is $\frac{6}{y_0}$.

At $B(12, 12)$, the slope of the tangent is $\frac{6}{12} = \frac{1}{2}$. This matches List-II (3).

(S) Maximum area of $\triangle ABC$ is $S(12)$.

$S(12) = \frac{3(12)^2}{8} - \frac{(12)^3}{48} = \frac{3 \times 144}{8} - \frac{1728}{48} = 3 \times 18 - 36 = 54 - 36 = 18$.

This matches List-II (4).

The final matching is:

(P) - (1) (Q) - (2) (R) - (3) (S) - (4)