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MCQMediumJEE 2024Basics: Distance, Section Formula, Locus

JEE Mathematics 2024 Question with Solution

If the locus of the point, whose distances from the point (2,1)(2, 1) and (1,3)(1, 3) are in the ratio 5:45:4, is ax2+by2+cxy+dx+ey+170=0ax^2 + by^2 + cxy + dx + ey + 170 = 0, then the value of a2+2b+3c+4d+ea^2 + 2b + 3c + 4d + e is equal to:

  • A

    55

  • B

    2727

  • C

    3737

  • D

    437437

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The distances of a point P(x,y)P(x,y) from (2,1)(2,1) and (1,3)(1,3) are in the ratio 5:45:4.

Find: The value of a2+2b+3c+4d+ea^2 + 2b + 3c + 4d + e when the locus is written as ax2+by2+cxy+dx+ey+170=0ax^2 + by^2 + cxy + dx + ey + 170 = 0.

Let P(x,y)P(x,y) be the moving point. Then

(x−2)2+(y−1)2(x−1)2+(y−3)2=2516\frac{(x - 2)^2 + (y - 1)^2}{(x - 1)^2 + (y - 3)^2} = \frac{25}{16}

Cross-multiplying and simplifying,

16[(x−2)2+(y−1)2]=25[(x−1)2+(y−3)2]16\left[(x - 2)^2 + (y - 1)^2\right] = 25\left[(x - 1)^2 + (y - 3)^2\right]

which gives

9x2+9y2+14x−118y+170=09x^2 + 9y^2 + 14x - 118y + 170 = 0

Comparing with ax2+by2+cxy+dx+ey+170=0ax^2 + by^2 + cxy + dx + ey + 170 = 0, we get

a=9,b=9,c=0,d=14,e=−118a = 9, \quad b = 9, \quad c = 0, \quad d = 14, \quad e = -118

Now,

a2+2b+3c+4d+e=81+18+0+56−118a^2 + 2b + 3c + 4d + e = 81 + 18 + 0 + 56 - 118 =155−118= 155 - 118 =37= 37

Therefore, the correct option is C.

Expanded Algebra

Given: d1d2=54\dfrac{d_1}{d_2} = \dfrac{5}{4} where

d1=(x−2)2+(y−1)2,d2=(x−1)2+(y−3)2d_1 = \sqrt{(x - 2)^2 + (y - 1)^2}, \qquad d_2 = \sqrt{(x - 1)^2 + (y - 3)^2}

Find: The value of a2+2b+3c+4d+ea^2 + 2b + 3c + 4d + e.

Squaring the ratio,

(x−2)2+(y−1)2(x−1)2+(y−3)2=2516\frac{(x - 2)^2 + (y - 1)^2}{(x - 1)^2 + (y - 3)^2} = \frac{25}{16}

So,

16[(x−2)2+(y−1)2]=25[(x−1)2+(y−3)2]16\left[(x - 2)^2 + (y - 1)^2\right] = 25\left[(x - 1)^2 + (y - 3)^2\right]

Expand both sides:

16[x2−4x+4+y2−2y+1]=25[x2−2x+1+y2−6y+9]16\left[x^2 - 4x + 4 + y^2 - 2y + 1\right] = 25\left[x^2 - 2x + 1 + y^2 - 6y + 9\right] 16x2−64x+16y2−32y+80=25x2−50x+25y2−150y+25016x^2 - 64x + 16y^2 - 32y + 80 = 25x^2 - 50x + 25y^2 - 150y + 250

Bringing all terms to one side,

−9x2−9y2−14x+118y−170=0-9x^2 - 9y^2 - 14x + 118y - 170 = 0

Multiplying by −1-1,

9x2+9y2+14x−118y+170=09x^2 + 9y^2 + 14x - 118y + 170 = 0

Thus,

a=9,b=9,c=0,d=14,e=−118a = 9, \quad b = 9, \quad c = 0, \quad d = 14, \quad e = -118

Hence,

a2+2b+3c+4d+e=92+2×9+3×0+4×14−118=37a^2 + 2b + 3c + 4d + e = 9^2 + 2 \times 9 + 3 \times 0 + 4 \times 14 - 118 = 37

Therefore, the value is 3737 and the correct option is C.

Common mistakes

  • Using the distance ratio directly without squaring it is incorrect because the distance formula contains square roots. First write d1d2=54\dfrac{d_1}{d_2} = \dfrac{5}{4} and then square both sides.

  • Missing the sign while moving terms to one side gives wrong coefficients for dd and ee. After expansion, carefully collect like terms before comparing with ax2+by2+cxy+dx+ey+170=0ax^2 + by^2 + cxy + dx + ey + 170 = 0.

  • Forgetting that the xyxy term is absent leads to taking c≠0c \neq 0. Here there is no mixed term, so c=0c = 0.

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