Let P ( 2 37, 6 7), Q, R, and S be four points on the ellipse 9x^… — Solution | JEE Mathematics

MCQMediumJEE 2023Conic Sections (Parabola, Ellipse, Hyperbola)

Let $P \left( \frac{2\sqrt{3}}{7}, \frac{6}{\sqrt{7}} \right), Q, R, \text{ and } S$ be four points on the ellipse $9x^2 + 4y^2 = 36$ . Let $PQ$ and $RS$ be mutually perpendicular and pass through the origin. If

\frac{1}{(PQ)^2} + \frac{1}{(RS)^2} = \frac{p}{q}

where $p$ and $q$ are coprime, then $p + q$ is equal to:

A
$143$
B
$137$
C
$157$
D
$147$

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: $P\left( \frac{2\sqrt{3}}{7}, \frac{6}{\sqrt{7}} \right)$ lies on the ellipse $9x^2 + 4y^2 = 36$ , and the chords $PQ$ and $RS$ pass through the origin and are mutually perpendicular.

Find: The value of $p+q$ if

\frac{1}{(PQ)^2} + \frac{1}{(RS)^2} = \frac{p}{q}

Let

R(2\cos\theta, 3\sin\theta)

represent the point such that $OP \perp OR$ .

To verify the perpendicularity condition, use the relation given in the solution:

3\sin\theta \times \frac{6}{\sqrt{7}} + 2\cos\theta \times \frac{2\sqrt{3}}{\sqrt{7}} = -1

Hence,

\tan\theta = -\frac{2}{3\sqrt{3}}

Using this, the possible coordinates of $R$ are

R\left( -\frac{6\sqrt{3}}{\sqrt{31}}, \frac{6}{\sqrt{31}} \right)

R\left( \frac{6\sqrt{3}}{\sqrt{31}}, -\frac{6}{\sqrt{31}} \right)

Now,

\frac{1}{(PQ)^2} + \frac{1}{(RS)^2} = \frac{1}{4} \left[ \frac{1}{(OP)^2} + \frac{1}{(OR)^2} \right]

Substituting the values used in the solution,

= \frac{1}{4} \left[ \frac{1}{\frac{48}{7}} + \frac{1}{\frac{144}{31}} \right]

= \frac{1}{4} \left[ \frac{7}{48} + \frac{31}{144} \right]

= \frac{1}{4} \left[ \frac{21+31}{144} \right]

= \frac{1}{4} \cdot \frac{52}{144} = \frac{13}{144}

Therefore,

p = 13, \quad q = 144

So,

p+q = 13+144 = 157

The solution working gives $157$ , which corresponds to option C. The solution's labels the correct option as D, but that disagrees with the computed value and the listed options.

Common mistakes

Treating $PQ$ and $RS$ as arbitrary segments instead of chords through the origin. Since each passes through the origin, the origin is the midpoint of each corresponding chord, so $PQ = 2\,OP$ and $RS = 2\,OR$ . Use this before forming reciprocal squares.
Using an incorrect parametrization of the ellipse. For $9x^2 + 4y^2 = 36$ , the correct parametric point is $\left(2\cos\theta, 3\sin\theta\right)$ , not $\left(3\cos\theta, 2\sin\theta\right)$ . Swapping semi-axes changes all distances.
Confusing perpendicular lines with perpendicular position vectors numerically. The condition must be imposed carefully using the slopes or the vector relation for $OP$ and $OR$ . An incorrect perpendicularity setup leads to the wrong value of $\tan\theta$ .

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Answer

Step-by-step solution

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