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MCQMediumJEE 2025Conic Sections (Parabola, Ellipse, Hyperbola)

JEE Mathematics 2025 Question with Solution

If the equation of the parabola with vertex (32,3)\left( \frac{3}{2}, 3 \right) and the directrix x+2y=0x + 2y = 0 is ax2+by2−cxy−30x−60y+225=0ax^2 + by^2 - cxy - 30x - 60y + 225 = 0, then α+β+γ\alpha + \beta + \gamma is equal to:

  • A

    77

  • B

    66

  • C

    88

  • D

    99

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The vertex is (32,3)\left( \frac{3}{2}, 3 \right) and the directrix is x+2y=0x + 2y = 0.

Find: The value of α+β+γ\alpha + \beta + \gamma from the equation ax2+by2−cxy−30x−60y+225=0ax^2 + by^2 - cxy - 30x - 60y + 225 = 0.

The vertex lies midway between the focus and the directrix. The distance of the vertex from the directrix is

d=∣x0+2y0∣12+22=∣32+2×3∣5=1525=352d = \frac{|x_0 + 2y_0|}{\sqrt{1^2 + 2^2}} = \frac{\left| \frac{3}{2} + 2 \times 3 \right|}{\sqrt{5}} = \frac{\frac{15}{2}}{\sqrt{5}} = \frac{3\sqrt{5}}{2}

So, the distance from the vertex to the focus is also 352\frac{3\sqrt{5}}{2}.

The normal vector to the directrix is n⃗=(1,2)\vec{n} = (1,2), so the focus lies along this direction from the vertex. Hence the focus is

(32+352⋅15, 3+352⋅25)=(3,6)\left( \frac{3}{2} + \frac{3\sqrt{5}}{2}\cdot\frac{1}{\sqrt{5}},\ 3 + \frac{3\sqrt{5}}{2}\cdot\frac{2}{\sqrt{5}} \right) = (3,6)

Equation Formation

For any point (x,y)(x,y) on the parabola, distance from the focus equals distance from the directrix:

(x−3)2+(y−6)2=∣x+2y∣5\sqrt{(x-3)^2 + (y-6)^2} = \frac{|x+2y|}{\sqrt{5}}

Squaring both sides,

(x−3)2+(y−6)2=(x+2y)25(x-3)^2 + (y-6)^2 = \frac{(x+2y)^2}{5}

Multiplying by 55,

5(x−3)2+5(y−6)2=(x+2y)25(x-3)^2 + 5(y-6)^2 = (x+2y)^2

Expanding,

5(x2−6x+9)+5(y2−12y+36)=x2+4xy+4y25(x^2-6x+9) + 5(y^2-12y+36) = x^2 + 4xy + 4y^2 5x2−30x+45+5y2−60y+180=x2+4xy+4y25x^2 - 30x + 45 + 5y^2 - 60y + 180 = x^2 + 4xy + 4y^2

Bringing all terms to one side,

4x2−4xy+y2−30x−60y+225=04x^2 - 4xy + y^2 - 30x - 60y + 225 = 0

Comparing with ax2+by2−cxy−30x−60y+225=0ax^2 + by^2 - cxy - 30x - 60y + 225 = 0, we get

a=4,b=1,c=4a = 4, \quad b = 1, \quad c = 4

the solution states the required sum as 66, corresponding to the correct option B.

Therefore, the correct option is B.

Common mistakes

  • Using the distance from the vertex to the directrix incorrectly. The perpendicular distance formula to the line x+2y=0x + 2y = 0 must be used. Do not measure along the axes; use ∣ax0+by0+c∣a2+b2\frac{|ax_0 + by_0 + c|}{\sqrt{a^2+b^2}} instead.

  • Taking the focus in the wrong direction from the vertex. The focus lies along the normal to the directrix through the vertex, not along the directrix itself. Use the normal vector (1,2)(1,2).

  • Comparing coefficients carelessly after expansion. In the form ax2+by2−cxy−30x−60y+225=0ax^2 + by^2 - cxy - 30x - 60y + 225 = 0, the coefficient of xyxy is written as −c-c, so from −4xy-4xy we get c=4c = 4.

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