If the equation of the parabola with vertex V(\frac32, 3) and the directrix x+2 y=0 is \alpha x^2+\beta y^2-\gamma x y-30 x-60 y+225=0 , then \alpha+\beta+\gamma is equal to :

Parabola PYQ — JEE Main 2025 · 24 Jan · Shift 2 · Q37 with Solution

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Correct answer: D

Given data

Vertex of parabola: $V\left(\frac{3}{2},3\right)$ Directrix: $x+2y=0$

We use the fact that for a parabola, the axis is perpendicular to the directrix and passes through the vertex.

Find the axis direction

Since the directrix is $x+2y=0,$ its normal vector is $(1,2).$ So the axis of the parabola is along the direction $(1,2)$ .

The distance from the vertex to the directrix is the focal length $a$ .

Distance of $V\left(\frac32,3\right)$ from the line $x+2y=0$ is $a=\frac{\left|\frac32+2\cdot 3\right|}{\sqrt{1^2+2^2}}=\frac{\left|\frac32+6\right|}{\sqrt5}=\frac{\frac{15}{2}}{\sqrt5}=\frac{15}{2\sqrt5}.$

So the focus lies from the vertex in the direction of $(1,2)$ , away from the directrix.

Unit vector along axis: $\hat{u}=\left(\frac1{\sqrt5},\frac2{\sqrt5}\right).$

Hence focus is

\left(\frac32,3\right)+\frac{15}{2\sqrt5}\left(\frac1{\sqrt5},\frac2{\sqrt5}\right) =\left(\frac32,3\right)+\left(\frac32,3\right)=(3,6).$$ --- 3. **Use definition of parabola** For any point $P(x,y)$ on the parabola, $$PF=\text{distance from }P\text{ to directrix}.$$ So, $$\sqrt{(x-3)^2+(y-6)^2}=\frac{|x+2y|}{\sqrt5}.$$ Squaring, $$5\big((x-3)^2+(y-6)^2\big)=(x+2y)^2.$$ --- 4. **Expand and simplify** First expand the left side: $$5\left[(x^2-6x+9)+(y^2-12y+36)\right] =5x^2+5y^2-30x-60y+225.$$ Right side: $$(x+2y)^2=x^2+4y^2+4xy.$$ Therefore, $$5x^2+5y^2-30x-60y+225=x^2+4y^2+4xy.$$ Bring all terms to one side: $$4x^2+y^2-4xy-30x-60y+225=0.$$ Compare with $$\alpha x^2+\beta y^2-\gamma xy-30x-60y+225=0.$$ Thus, $$\alpha=4,\quad \beta=1,\quad \gamma=4.$$ So, $$\alpha+\beta+\gamma=4+1+4=9.$$ --- 5. **Check options** The correct option is: $$\boxed{9}$$ which is **Option D**.

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View written solutionFree

Correct answer: D

Given data

Vertex of parabola: $V\left(\frac{3}{2},3\right)$ Directrix: $x+2y=0$

We use the fact that for a parabola, the axis is perpendicular to the directrix and passes through the vertex.

Find the axis direction

Since the directrix is $x+2y=0,$ its normal vector is $(1,2).$ So the axis of the parabola is along the direction $(1,2)$ .

The distance from the vertex to the directrix is the focal length $a$ .

So the focus lies from the vertex in the direction of $(1,2)$ , away from the directrix.

Unit vector along axis: $\hat{u}=\left(\frac1{\sqrt5},\frac2{\sqrt5}\right).$

Hence focus is

\left(\frac32,3\right)+\frac{15}{2\sqrt5}\left(\frac1{\sqrt5},\frac2{\sqrt5}\right) =\left(\frac32,3\right)+\left(\frac32,3\right)=(3,6).$$ --- 3. **Use definition of parabola** For any point $P(x,y)$ on the parabola, $$PF=\text{distance from }P\text{ to directrix}.$$ So, $$\sqrt{(x-3)^2+(y-6)^2}=\frac{|x+2y|}{\sqrt5}.$$ Squaring, $$5\big((x-3)^2+(y-6)^2\big)=(x+2y)^2.$$ --- 4. **Expand and simplify** First expand the left side: $$5\left[(x^2-6x+9)+(y^2-12y+36)\right] =5x^2+5y^2-30x-60y+225.$$ Right side: $$(x+2y)^2=x^2+4y^2+4xy.$$ Therefore, $$5x^2+5y^2-30x-60y+225=x^2+4y^2+4xy.$$ Bring all terms to one side: $$4x^2+y^2-4xy-30x-60y+225=0.$$ Compare with $$\alpha x^2+\beta y^2-\gamma xy-30x-60y+225=0.$$ Thus, $$\alpha=4,\quad \beta=1,\quad \gamma=4.$$ So, $$\alpha+\beta+\gamma=4+1+4=9.$$ --- 5. **Check options** The correct option is: $$\boxed{9}$$ which is **Option D**.

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