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Ellipse question

2021 · 25 Feb · Shift 2 · Q38
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  5. /2021 · 25 Feb · Shift 2 · Q38

Ellipse question

2021 · 25 Feb · Shift 2 · Q38

JEE MainMathematicsEllipseMCQ+4 / −1
If the curve x2 + 2y2 = 2 intersects the line x + y = 1 at two points P and Q, then the angle subtended by the line segment PQ at the origin is :
  1. A
    π2−tan⁡−1(14){\pi \over 2} - {\tan ^{ - 1}}\left( {{1 \over 4}} \right)2π​−tan−1(41​)
  2. B
    π2+tan⁡−1(13){\pi \over 2} + {\tan ^{ - 1}}\left( {{1 \over 3}} \right)2π​+tan−1(31​)
  3. C
    π2−tan⁡−1(13){\pi \over 2} - {\tan ^{ - 1}}\left( {{1 \over 3}} \right)2π​−tan−1(31​)
  4. D
    π2+tan⁡−1(14){\pi \over 2} + {\tan ^{ - 1}}\left( {{1 \over 4}} \right)2π​+tan−1(41​)
View written solutionFree

Correct answer: D

  1. Given curves

We need the points of intersection of x2+2y2=2x^2+2y^2=2x2+2y2=2 and x+y=1.x+y=1.x+y=1.

Let the intersection points be PPP and QQQ.

  1. Find the intersection points

From the line, x=1−y.x=1-y.x=1−y. Substitute into the ellipse: (1−y)2+2y2=2.(1-y)^2+2y^2=2.(1−y)2+2y2=2.

Expanding, 1−2y+y2+2y2=21-2y+y^2+2y^2=21−2y+y2+2y2=2 3y2−2y−1=0.3y^2-2y-1=0.3y2−2y−1=0.

Solve: 3y2−2y−1=(3y+1)(y−1)=0.3y^2-2y-1=(3y+1)(y-1)=0.3y2−2y−1=(3y+1)(y−1)=0. So, y=1ory=−13.y=1 \quad \text{or} \quad y=-\frac13.y=1ory=−31​.

Corresponding xxx values from x=1−yx=1-yx=1−y:

  • If y=1y=1y=1, then x=0x=0x=0
  • If y=−13y=-\frac13y=−31​, then x=43x=\frac43x=34​

Hence, P=(0,1),Q=(43,−13).P=(0,1), \qquad Q=\left(\frac43,-\frac13\right).P=(0,1),Q=(34​,−31​).

  1. Angle subtended by PQPQPQ at the origin

This is the angle between the vectors OP⃗=(0,1),OQ⃗=(43,−13).\vec{OP}=(0,1), \qquad \vec{OQ}=\left(\frac43,-\frac13\right).OP=(0,1),OQ​=(34​,−31​).

Let the required angle be θ\thetaθ.

Using the dot product, cos⁡θ=OP⃗⋅OQ⃗∣OP∣ ∣OQ∣.\cos\theta=\frac{\vec{OP}\cdot\vec{OQ}}{|OP|\,|OQ|}.cosθ=∣OP∣∣OQ∣OP⋅OQ​​.

Now, OP⃗⋅OQ⃗=0⋅43+1⋅(−13)=−13.\vec{OP}\cdot\vec{OQ}=0\cdot \frac43+1\cdot\left(-\frac13\right)=-\frac13.OP⋅OQ​=0⋅34​+1⋅(−31​)=−31​.

Also, ∣OP∣=1,|OP|=1,∣OP∣=1, ∣OQ∣=(43)2+(−13)2=169+19=173.|OQ|=\sqrt{\left(\frac43\right)^2+\left(-\frac13\right)^2}=\sqrt{\frac{16}{9}+\frac{1}{9}}=\frac{\sqrt{17}}{3}.∣OQ∣=(34​)2+(−31​)2​=916​+91​​=317​​.

Therefore, cos⁡θ=−1/317/3=−117.\cos\theta=\frac{-1/3}{\sqrt{17}/3}=-\frac{1}{\sqrt{17}}.cosθ=17​/3−1/3​=−17​1​.

So, θ=cos⁡−1(−117).\theta=\cos^{-1}\left(-\frac{1}{\sqrt{17}}\right).θ=cos−1(−17​1​).

  1. Convert to the given form

Let α=tan⁡−1(4).\alpha=\tan^{-1}(4).α=tan−1(4). Then in a right triangle, cos⁡α=117,sin⁡α=417.\cos\alpha=\frac{1}{\sqrt{17}}, \qquad \sin\alpha=\frac{4}{\sqrt{17}}.cosα=17​1​,sinα=17​4​.

Hence, cos⁡θ=−117=−cos⁡α.\cos\theta=-\frac{1}{\sqrt{17}}=-\cos\alpha.cosθ=−17​1​=−cosα. So, θ=π−α.\theta=\pi-\alpha.θ=π−α.

Now use tan⁡−1(4)=π2−tan⁡−1(14).\tan^{-1}(4)=\frac{\pi}{2}-\tan^{-1}\left(\frac14\right).tan−1(4)=2π​−tan−1(41​). Therefore,

=\frac{\pi}{2}+\tan^{-1}\left(\frac14\right).$$ Thus the required angle is $$\boxed{\frac{\pi}{2}+\tan^{-1}\left(\frac14\right)}.$$ --- 5. **Check options** This matches **Option D**. --- 6. **Compare with stored correct answer** Stored correct answer: **D** Our derived answer: **D** So they agree.
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