JEE Mathematics 2023 Question with Solution

Given: The ellipse is $x^2 + 9y^2 = 9$. Its major axis lies along the $x$-axis with endpoints $(-3,0)$ and $(3,0)$, so the circle having this major axis as diameter is

$$x^2 + y^2 = 9$$

Point $A=(3,0)$ and point $B=(0,1)$.

Find: $m-n$, where the area of triangle $AOP$ is $\frac{m}{n}$.

The line through $A(3,0)$ and $B(0,1)$ is

$$\frac{x}{3} + \frac{y}{1} = 1$$

which gives

$$x + 3y = 3$$

or

$$x = 3 - 3y$$

Substitute this in the circle equation:

$$(3 - 3y)^2 + y^2 = 9$$ $$9(1 + y^2 - 2y) + y^2 = 9$$ $$10y^2 - 18y = 0$$ $$y(10y - 18) = 0$$

So the second intersection point is obtained from

$$y = \frac{9}{5}$$

Now,

$$x = 3 - 3\left(\frac{9}{5}\right) = 3\left(1 - \frac{9}{5}\right) = -\frac{12}{5}$$

Hence,

$$P\left(-\frac{12}{5}, \frac{9}{5}\right)$$

Area of triangle $AOP$ is

$$\frac{1}{2} \times \text{Base}(OA) \times \text{Height}$$

Here, $OA = 3$ and the perpendicular height from $P$ to the $x$-axis is $\frac{9}{5}$. Therefore,

$$\text{Area} = \frac{1}{2} \times 3 \times \frac{9}{5} = \frac{27}{10}$$

So,

$$m = 27, \quad n = 10$$

Thus,

$$m-n = 27-10 = 17$$

Therefore, the correct option is D. The solution labels it as Option C, but the computed value $17$ matches option D in the given options.