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Parabola question

2015 · Shift 1 · Q26
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Parabola question

2015 · Shift 1 · Q26

JEE AdvancedMathematicsParabolaNumerical+4 / −1
Let the curve CCC be the mirror image of the parabola y2=4x{y^2} = 4xy2=4x with respect to the line x+y+4=0x+y+4=0x+y+4=0. If AAA and BBB are the points of intersection of CCC with the line y=−5y=-5y=−5, then the distance between AAA and BBB is
Numerical answer
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Correct answer: 4

  1. Given parabola and reflection line

The parabola is y2=4xy^2=4xy2=4x and it is reflected about the line x+y+4=0.x+y+4=0.x+y+4=0.

We need the intersection points of the reflected curve CCC with the line y=−5.y=-5.y=−5. Then we must find the distance between those two points.

  1. Use the reflection property smartly

Instead of finding the full reflected parabola, reflect the line y=−5y=-5y=−5 in the same mirror line.
Then its intersections with the original parabola will correspond to the reflected parabola's intersections with y=−5y=-5y=−5.

So we reflect the line y=−5y=-5y=−5 about x+y+4=0.x+y+4=0.x+y+4=0.

  1. Reflect points on the line y=−5y=-5y=−5

Take a general point on y=−5y=-5y=−5: (t,−5).(t,-5).(t,−5).

Reflection of point (x1,y1)(x_1,y_1)(x1​,y1​) about line ax+by+c=0ax+by+c=0ax+by+c=0 is given by x′=x1−2a(ax1+by1+c)a2+b2,x'=x_1-\frac{2a(ax_1+by_1+c)}{a^2+b^2},x′=x1​−a2+b22a(ax1​+by1​+c)​, y′=y1−2b(ax1+by1+c)a2+b2.y'=y_1-\frac{2b(ax_1+by_1+c)}{a^2+b^2}.y′=y1​−a2+b22b(ax1​+by1​+c)​.

Here, a=1,b=1,c=4,a=1,\quad b=1,\quad c=4,a=1,b=1,c=4, and (x1,y1)=(t,−5)(x_1,y_1)=(t,-5)(x1​,y1​)=(t,−5).

Compute: ax1+by1+c=t−5+4=t−1.ax_1+by_1+c=t-5+4=t-1.ax1​+by1​+c=t−5+4=t−1.

Thus x′=t−2(t−1)2=t−(t−1)=1,x'=t-\frac{2(t-1)}{2}=t-(t-1)=1,x′=t−22(t−1)​=t−(t−1)=1, y′=−5−2(t−1)2=−5−(t−1)=−t−4.y'=-5-\frac{2(t-1)}{2}=-5-(t-1)=-t-4.y′=−5−22(t−1)​=−5−(t−1)=−t−4.

So the reflected point is (1,−t−4).(1,-t-4).(1,−t−4).

As ttt varies, all reflected points satisfy x′=1.x'=1.x′=1.

Hence the mirror image of the line y=−5y=-5y=−5 is x=1.x=1.x=1.

  1. Intersect the original parabola with the reflected line

Now intersect y2=4xy^2=4xy2=4x with x=1.x=1.x=1.

Substitute x=1x=1x=1: y2=4  ⟹  y=±2.y^2=4\implies y=\pm 2.y2=4⟹y=±2.

So the intersection points on the original parabola are (1,2)and(1,−2).(1,2)\quad \text{and} \quad (1,-2).(1,2)and(1,−2).

These correspond, after reflection, to the required points AAA and BBB on the reflected parabola CCC lying on y=−5y=-5y=−5.

  1. Reflect these two points to get AAA and BBB

Reflect (1,2)(1,2)(1,2) about x+y+4=0x+y+4=0x+y+4=0:

First compute 1+2+4=7.1+2+4=7.1+2+4=7. Then x′=1−2⋅1⋅72=1−7=−6,x'=1-\frac{2\cdot 1\cdot 7}{2}=1-7=-6,x′=1−22⋅1⋅7​=1−7=−6, y′=2−2⋅1⋅72=2−7=−5.y'=2-\frac{2\cdot 1\cdot 7}{2}=2-7=-5.y′=2−22⋅1⋅7​=2−7=−5. So one point is A=(−6,−5).A=(-6,-5).A=(−6,−5).

Reflect (1,−2)(1,-2)(1,−2):

First compute 1−2+4=3.1-2+4=3.1−2+4=3. Then x′=1−2⋅1⋅32=1−3=−2,x'=1-\frac{2\cdot 1\cdot 3}{2}=1-3=-2,x′=1−22⋅1⋅3​=1−3=−2, y′=−2−2⋅1⋅32=−2−3=−5.y'=-2-\frac{2\cdot 1\cdot 3}{2}=-2-3=-5.y′=−2−22⋅1⋅3​=−2−3=−5. So the other point is B=(−2,−5).B=(-2,-5).B=(−2,−5).

  1. Distance between AAA and BBB

Since both points lie on y=−5y=-5y=−5, the distance is just the difference of xxx-coordinates: AB=∣−2−(−6)∣=4.AB=|-2-(-6)|=4.AB=∣−2−(−6)∣=4.

  1. Final answer

The required distance is 4.\boxed{4}.4​.

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