Let P(\frac2 \sqrt3\sqrt7, \frac6\sqrt7), Q, R and S be four points on the ellipse 9 x^2+4 y^2=36 . Let PQ and RS be mutually perpendicular and pass through the origin. If \frac1(P Q)^2+\frac1(R S)^2=\fracpq , where p and q are coprime, then p+q is equal to :

Ellipse PYQ — JEE Main 2023 · 12 Apr · Shift 1 · Q31 with Solution

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Correct answer: D

Write the ellipse in standard form

Given $9x^2+4y^2=36$ Divide by $36$ : $\frac{x^2}{4}+\frac{y^2}{9}=1$ So the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with $a^2=4,\ b^2=9$ .

Find the line through the origin and point $P$

Given $P\left(\frac{2\sqrt3}{\sqrt7},\frac{6}{\sqrt7}\right)$ The slope of $OP$ is $m_1=\frac{\frac{6}{\sqrt7}}{\frac{2\sqrt3}{\sqrt7}}=\frac{6}{2\sqrt3}=\sqrt3$ Hence chord $PQ$ lies on the line $y=\sqrt3\,x$ Since this line passes through the origin and cuts the ellipse at two points, $PQ$ is the chord along this line.

Find the length of chord $PQ$

Substitute $y=\sqrt3 x$ into the ellipse: $9x^2+4(3x^2)=36$ $9x^2+12x^2=36$ $21x^2=36$ $x^2=\frac{12}{7}$ So the intersection points are at $x=\pm \frac{2\sqrt3}{\sqrt7}$ which matches the given point $P$ .

Thus the two endpoints are opposite points through the origin, so $PQ=2\cdot OP$ Now

=\frac{12}{7}+\frac{36}{7}=\frac{48}{7}$$ Hence $$PQ^2=4\cdot OP^2=4\cdot \frac{48}{7}=\frac{192}{7}$$ Therefore $$\frac{1}{(PQ)^2}=\frac{7}{192}$$ --- 4. **Find the perpendicular chord through the origin** Since $RS$ is perpendicular to $PQ$, its slope is $$m_2=-\frac{1}{\sqrt3}$$ So the line is $$y=-\frac{x}{\sqrt3}$$ Substitute into the ellipse: $$9x^2+4\left(\frac{x^2}{3}\right)=36$$ $$9x^2+\frac{4x^2}{3}=36$$ Multiply by $3$: $$27x^2+4x^2=108$$ $$31x^2=108$$ $$x^2=\frac{108}{31}$$ Then $$y^2=\frac{x^2}{3}=\frac{36}{31}$$ The distance from origin to either intersection point is $$OR^2=x^2+y^2=\frac{108}{31}+\frac{36}{31}=\frac{144}{31}$$ Thus $$RS^2=4\cdot OR^2=4\cdot \frac{144}{31}=\frac{576}{31}$$ So $$\frac{1}{(RS)^2}=\frac{31}{576}$$ --- 5. **Add the reciprocals** $$\frac{1}{(PQ)^2}+\frac{1}{(RS)^2}=\frac{7}{192}+\frac{31}{576}$$ Take LCM $576$: $$\frac{7}{192}=\frac{21}{576}$$ Therefore $$\frac{21}{576}+\frac{31}{576}=\frac{52}{576}=\frac{13}{144}$$ So $$\frac{p}{q}=\frac{13}{144}$$ with coprime $p=13, q=144$. Hence $$p+q=13+144=157$$ --- 6. **Check the options** The correct option is: $$\boxed{157}$$ which is **Option D**.

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View written solutionFree

Correct answer: D

Write the ellipse in standard form

Given $9x^2+4y^2=36$ Divide by $36$ : $\frac{x^2}{4}+\frac{y^2}{9}=1$ So the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with $a^2=4,\ b^2=9$ .

Find the line through the origin and point $P$

Find the length of chord $PQ$

Thus the two endpoints are opposite points through the origin, so $PQ=2\cdot OP$ Now

=\frac{12}{7}+\frac{36}{7}=\frac{48}{7}$$ Hence $$PQ^2=4\cdot OP^2=4\cdot \frac{48}{7}=\frac{192}{7}$$ Therefore $$\frac{1}{(PQ)^2}=\frac{7}{192}$$ --- 4. **Find the perpendicular chord through the origin** Since $RS$ is perpendicular to $PQ$, its slope is $$m_2=-\frac{1}{\sqrt3}$$ So the line is $$y=-\frac{x}{\sqrt3}$$ Substitute into the ellipse: $$9x^2+4\left(\frac{x^2}{3}\right)=36$$ $$9x^2+\frac{4x^2}{3}=36$$ Multiply by $3$: $$27x^2+4x^2=108$$ $$31x^2=108$$ $$x^2=\frac{108}{31}$$ Then $$y^2=\frac{x^2}{3}=\frac{36}{31}$$ The distance from origin to either intersection point is $$OR^2=x^2+y^2=\frac{108}{31}+\frac{36}{31}=\frac{144}{31}$$ Thus $$RS^2=4\cdot OR^2=4\cdot \frac{144}{31}=\frac{576}{31}$$ So $$\frac{1}{(RS)^2}=\frac{31}{576}$$ --- 5. **Add the reciprocals** $$\frac{1}{(PQ)^2}+\frac{1}{(RS)^2}=\frac{7}{192}+\frac{31}{576}$$ Take LCM $576$: $$\frac{7}{192}=\frac{21}{576}$$ Therefore $$\frac{21}{576}+\frac{31}{576}=\frac{52}{576}=\frac{13}{144}$$ So $$\frac{p}{q}=\frac{13}{144}$$ with coprime $p=13, q=144$. Hence $$p+q=13+144=157$$ --- 6. **Check the options** The correct option is: $$\boxed{157}$$ which is **Option D**.

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