Let y^2=12 x be the parabola and S be its focus. Let P Q be a focal chord of the parabola such that (S P)(S Q)=\frac1474 . Let C be the circle described taking P Q as a diameter. If the equation of a circle C is 64 x^2+64 y^2-\alpha x-64 \sqrt3 y=\beta , then \beta-\alpha is equal to \qquad .

Parabola PYQ — JEE Main 2025 · 29 Jan · Shift 2 · Q47 with Solution

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Correct answer: 1328

Given parabola and its focus

The parabola is $y^2=12x.$ Comparing with the standard form $y^2=4ax$ , we get $4a=12 \implies a=3.$ So the focus is $S=(3,0).$

Parametric points on the parabola

A general point on $y^2=4ax$ is $(at^2,2at).$ Here $a=3$ , so a point is $(3t^2,6t).$

Let the endpoints of the focal chord be $P=(3t_1^2,6t_1), \qquad Q=(3t_2^2,6t_2).$

For a focal chord of a parabola, the parameters satisfy $t_1t_2=-1.$ So let $t_1=t, \qquad t_2=-\frac1t.$

Thus, $P=(3t^2,6t), \qquad Q=\left(\frac{3}{t^2},-\frac{6}{t}\right).$

Use the condition $(SP)(SQ)=\dfrac{147}{4}$

For the parabola $y^2=4ax$ , distance from focus to point $(at^2,2at)$ is $SP=a(1+t^2).$ Here $a=3$ , hence $SP=3(1+t^2).$

Similarly, for parameter $-1/t$ , $SQ=3\left(1+\frac1{t^2}\right).$

Therefore

=9\cdot \frac{(1+t^2)^2}{t^2}.$$ Given that $$9\cdot \frac{(1+t^2)^2}{t^2}=\frac{147}{4}.$$ Divide by $9$: $$\frac{(1+t^2)^2}{t^2}=\frac{147}{36}=\frac{49}{12}.$$ So $$(1+t^2)^2=\frac{49}{12}t^2.$$ Let $$u=t+\frac1t.$$ Then $$u^2=t^2+2+\frac1{t^2}=\frac{(1+t^2)^2}{t^2}.$$ Hence $$u^2=\frac{49}{12} \implies u=\pm \frac{7}{2\sqrt3}.$$ --- 4. **Equation of the circle with diameter $PQ$** If $P(x_1,y_1)$ and $Q(x_2,y_2)$ are endpoints of a diameter, then the circle is $$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0.$$ Here, $$x_1=3t^2,\quad y_1=6t,\quad x_2=\frac{3}{t^2},\quad y_2=-\frac{6}{t}.$$ So, $$x^2-(x_1+x_2)x+x_1x_2+y^2-(y_1+y_2)y+y_1y_2=0.$$ Now compute: - $$x_1+x_2=3\left(t^2+\frac1{t^2}\right),$$ - $$x_1x_2=9,$$ - $$y_1+y_2=6\left(t-\frac1t\right),$$ - $$y_1y_2=-36.$$ Hence, $$x^2+y^2-3\left(t^2+\frac1{t^2}\right)x-6\left(t-\frac1t\right)y+(9-36)=0,$$ that is, $$x^2+y^2-3\left(t^2+\frac1{t^2}\right)x-6\left(t-\frac1t\right)y-27=0.$$ Multiplying by $64$, $$64x^2+64y^2-192\left(t^2+\frac1{t^2}\right)x-384\left(t-\frac1t\right)y-1728=0.$$ Rewrite as $$64x^2+64y^2-\alpha x-64\sqrt3\,y=\beta.$$ Therefore, $$\alpha=192\left(t^2+\frac1{t^2}\right),$$ $$64\sqrt3=384\left(t-\frac1t\right) \implies t-\frac1t=\frac{\sqrt3}{6},$$ up to sign. Since the given coefficient is $-64\sqrt3 y$, we need $$384\left(t-\frac1t\right)=64\sqrt3 \implies t-\frac1t=\frac{\sqrt3}{6}.$$ Also, $$\beta=1728.$$ --- 5. **Find $t^2+\dfrac1{t^2}$** From $$t-\frac1t=\frac{\sqrt3}{6},$$ we get $$\left(t-\frac1t\right)^2=t^2+\frac1{t^2}-2=\frac{3}{36}=\frac1{12}.$$ Thus, $$t^2+\frac1{t^2}=2+\frac1{12}=\frac{25}{12}.$$ So $$\alpha=192\cdot \frac{25}{12}=16\cdot 25=400.$$ Therefore, $$\beta-\alpha=1728-400=1328.$$ --- 6. **Final answer** $$\boxed{1328}$$ This matches the stored correct answer.

More from Parabola

View written solutionFree

Correct answer: 1328

Given parabola and its focus

The parabola is $y^2=12x.$ Comparing with the standard form $y^2=4ax$ , we get $4a=12 \implies a=3.$ So the focus is $S=(3,0).$

Parametric points on the parabola

A general point on $y^2=4ax$ is $(at^2,2at).$ Here $a=3$ , so a point is $(3t^2,6t).$

Let the endpoints of the focal chord be $P=(3t_1^2,6t_1), \qquad Q=(3t_2^2,6t_2).$

For a focal chord of a parabola, the parameters satisfy $t_1t_2=-1.$ So let $t_1=t, \qquad t_2=-\frac1t.$

Thus, $P=(3t^2,6t), \qquad Q=\left(\frac{3}{t^2},-\frac{6}{t}\right).$

Use the condition $(SP)(SQ)=\dfrac{147}{4}$

For the parabola $y^2=4ax$ , distance from focus to point $(at^2,2at)$ is $SP=a(1+t^2).$ Here $a=3$ , hence $SP=3(1+t^2).$

Similarly, for parameter $-1/t$ , $SQ=3\left(1+\frac1{t^2}\right).$

Therefore

=9\cdot \frac{(1+t^2)^2}{t^2}.$$ Given that $$9\cdot \frac{(1+t^2)^2}{t^2}=\frac{147}{4}.$$ Divide by $9$: $$\frac{(1+t^2)^2}{t^2}=\frac{147}{36}=\frac{49}{12}.$$ So $$(1+t^2)^2=\frac{49}{12}t^2.$$ Let $$u=t+\frac1t.$$ Then $$u^2=t^2+2+\frac1{t^2}=\frac{(1+t^2)^2}{t^2}.$$ Hence $$u^2=\frac{49}{12} \implies u=\pm \frac{7}{2\sqrt3}.$$ --- 4. **Equation of the circle with diameter $PQ$** If $P(x_1,y_1)$ and $Q(x_2,y_2)$ are endpoints of a diameter, then the circle is $$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0.$$ Here, $$x_1=3t^2,\quad y_1=6t,\quad x_2=\frac{3}{t^2},\quad y_2=-\frac{6}{t}.$$ So, $$x^2-(x_1+x_2)x+x_1x_2+y^2-(y_1+y_2)y+y_1y_2=0.$$ Now compute: - $$x_1+x_2=3\left(t^2+\frac1{t^2}\right),$$ - $$x_1x_2=9,$$ - $$y_1+y_2=6\left(t-\frac1t\right),$$ - $$y_1y_2=-36.$$ Hence, $$x^2+y^2-3\left(t^2+\frac1{t^2}\right)x-6\left(t-\frac1t\right)y+(9-36)=0,$$ that is, $$x^2+y^2-3\left(t^2+\frac1{t^2}\right)x-6\left(t-\frac1t\right)y-27=0.$$ Multiplying by $64$, $$64x^2+64y^2-192\left(t^2+\frac1{t^2}\right)x-384\left(t-\frac1t\right)y-1728=0.$$ Rewrite as $$64x^2+64y^2-\alpha x-64\sqrt3\,y=\beta.$$ Therefore, $$\alpha=192\left(t^2+\frac1{t^2}\right),$$ $$64\sqrt3=384\left(t-\frac1t\right) \implies t-\frac1t=\frac{\sqrt3}{6},$$ up to sign. Since the given coefficient is $-64\sqrt3 y$, we need $$384\left(t-\frac1t\right)=64\sqrt3 \implies t-\frac1t=\frac{\sqrt3}{6}.$$ Also, $$\beta=1728.$$ --- 5. **Find $t^2+\dfrac1{t^2}$** From $$t-\frac1t=\frac{\sqrt3}{6},$$ we get $$\left(t-\frac1t\right)^2=t^2+\frac1{t^2}-2=\frac{3}{36}=\frac1{12}.$$ Thus, $$t^2+\frac1{t^2}=2+\frac1{12}=\frac{25}{12}.$$ So $$\alpha=192\cdot \frac{25}{12}=16\cdot 25=400.$$ Therefore, $$\beta-\alpha=1728-400=1328.$$ --- 6. **Final answer** $$\boxed{1328}$$ This matches the stored correct answer.

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