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Parabola question

2025 · 2 Apr · Shift 1 · Q31
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Parabola question

2025 · 2 Apr · Shift 1 · Q31

JEE MainMathematicsParabolaMCQ+4 / −1
Let the focal chord PQ of the parabola y2=4xy^2=4 xy2=4x make an angle of 60∘60^{\circ}60∘ with the positive xxx axis, where P lies in the first quadrant. If the circle, whose one diameter is PS, S being the focus of the parabola, touches the yyy-axis at the point (0,α)(0, \alpha)(0,α), then 5α25 \alpha^25α2 is equal to:
  1. A
    15
  2. B
    25
  3. C
    20
  4. D
    30
View written solutionFree

Correct answer: A

  1. Parabola and focus

For the parabola y2=4x,y^2=4x,y2=4x, we have a=1a=1a=1, so its focus is S=(1,0).S=(1,0).S=(1,0).

  1. Equation of the focal chord

A focal chord passes through the focus and makes an angle 60∘60^\circ60∘ with the positive xxx-axis. Hence its slope is m=tan⁡60∘=3.m=\tan 60^\circ=\sqrt{3}.m=tan60∘=3​. So the chord line is y=3(x−1).y=\sqrt{3}(x-1).y=3​(x−1).

  1. Find the intersection points with the parabola

Substitute y=3(x−1)y=\sqrt{3}(x-1)y=3​(x−1) into y2=4xy^2=4xy2=4x: 3(x−1)2=4x.3(x-1)^2=4x.3(x−1)2=4x. Expanding, 3(x2−2x+1)=4x3(x^2-2x+1)=4x3(x2−2x+1)=4x 3x2−6x+3=4x3x^2-6x+3=4x3x2−6x+3=4x 3x2−10x+3=0.3x^2-10x+3=0.3x2−10x+3=0.

Solving, x=10±100−366=10±86.x=\frac{10\pm\sqrt{100-36}}{6}=\frac{10\pm 8}{6}.x=610±100−36​​=610±8​. Thus, x=3orx=13.x=3 \quad \text{or} \quad x=\frac13.x=3orx=31​.

Now compute corresponding yyy values:

  • For x=3x=3x=3, y=3(3−1)=23.y=\sqrt{3}(3-1)=2\sqrt{3}.y=3​(3−1)=23​.
  • For x=13x=\frac13x=31​, y=3(13−1)=−233.y=\sqrt{3}\left(\frac13-1\right)=-\frac{2\sqrt{3}}{3}.y=3​(31​−1)=−323​​.

Since PPP lies in the first quadrant, P=(3,23).P=(3,2\sqrt{3}).P=(3,23​).

  1. Circle with diameter PSPSPS

Endpoints of diameter are P=(3,23),S=(1,0).P=(3,2\sqrt{3}), \qquad S=(1,0).P=(3,23​),S=(1,0).

Center of the circle is the midpoint: C=(3+12,23+02)=(2,3).C=\left(\frac{3+1}{2},\frac{2\sqrt{3}+0}{2}\right)=(2,\sqrt{3}).C=(23+1​,223​+0​)=(2,3​).

Radius is half of PSPSPS:

\sqrt{4+12}=4,$$ so $$r=2.$$ Thus the circle is $$(x-2)^2+(y-\sqrt{3})^2=4.$$ 5. **Point where the circle touches the $y$-axis** The $y$-axis is the line $x=0$. For tangency, the distance from center to the $y$-axis must equal the radius. Distance of $C=(2,\sqrt{3})$ from the $y$-axis is $2$, which equals the radius $2$, so the circle is indeed tangent to the $y$-axis. The point of contact is the foot of the perpendicular from the center to the $y$-axis: $$(0,\sqrt{3}).$$ Hence, $$\alpha=\sqrt{3}.$$ 6. **Compute $5\alpha^2$** $$5\alpha^2=5(\sqrt{3})^2=5\cdot 3=15.$$ 7. **Option check** $$15$$ corresponds to **Option A**. Therefore, the correct answer is **A**.
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