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QJEEM23Mab217: JEE Main 2023 Mathematics - Class 11 JEEM23Mab217 Solution: Let the ellipse intersect the positive x and y-axes at the…

Let the ellipse E:x2+9y2=9E:{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 mn{m \over n} , where m and n are coprime, then m−nm - n is equal to :

Multiple Choice Options

  • A 15
  • B 16
  • C 17
  • D 18

Step-by-Step Verified Solution

The given equation of the ellipse is

amp;x2+9y2=9 ..........(i)amp;⇒x29+y21=1\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 ABA B is

x+3y=3 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+3y=3x+3 y=3 and circle

amp;x2+y2=9amp;∴(3−3y)2+y2=9amp;⇒9−18y+9y2+y2=9amp;⇒−18y+10y2=0amp;⇒18y=10y2amp;⇒y=0,95\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,PA, P and OO makes a △APO\triangle A P O ,

whose area is 12×\frac{1}{2} \times Base ×\times Height

amp; Area =12×3×95=2710=mn    [Given]amp;∴m=27,n=10amp;⇒m−n=27−10=17\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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