JEE MAIN - Mathematics (2023 - 10th April Morning Shift - No. 3)

Let the ellipse $$E:{x^2} + 9{y^2} = 9$$ intersect the positive x and y-axes at the points A and B respectively. Let the major axis of E be a diameter of the circle C. Let the line passing through A and B meet the circle C at the point P. If the area of the triangle with vertices A, P and the origin O is $${m \over n}$$, where m and n are coprime, then $$m - n$$ is equal to :
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17
18

Explanation

The given equation of the ellipse is

$$ \begin{aligned} & x^2+9 y^2=9 ~..........(i)\\\\ & \Rightarrow \frac{x^2}{9}+\frac{y^2}{1}=1 \end{aligned} $$

Now, equation of line $A B$ is

$$ x+3 y=3 ~$$...........(ii)

JEE Main 2023 (Online) 10th April Morning Shift Mathematics - Ellipse Question 18 English Explanation

Now, point of intersection of line $x+3 y=3$ and circle

$$ \begin{array}{lr} &x^2+y^2=9 \\\\ &\therefore (3-3 y)^2+y^2=9 \\\\ &\Rightarrow 9-18 y+9 y^2+y^2=9 \\\\ &\Rightarrow -18 y+10 y^2=0 \\\\ &\Rightarrow 18 y=10 y^2 \\\\ &\Rightarrow y=0, \frac{9}{5} \end{array} $$

Now, from figure, we clearly see that vertices $A, P$ and $O$ makes a $\triangle A P O$,

whose area is $\frac{1}{2} \times$ Base $\times$ Height

$$ \begin{aligned} & \text { Area }=\frac{1}{2} \times 3 \times \frac{9}{5}=\frac{27}{10}=\frac{m}{n} ~~~~[Given]\\\\ & \therefore m=27, n=10 \\\\ & \Rightarrow m-n=27-10=17 \end{aligned} $$

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