Consider the circle C: x^2+y^2=4 and the parabola P: y^2=8 x . If the set of all values of \alpha , for which three chords of the circle C on three distinct lines passing through the point (\alpha, 0) are bisected by the parabola P is the interval (p, q) , then (2 q-p)^2 is equal to \underline\hspace2cm .

Parabola PYQ — JEE Main 2024 · 9 Apr · Shift 2 · Q55 with Solution

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Correct answer: 80

Interpret the condition

A chord of the circle $C:x^2+y^2=4$ lying on a line through $(\alpha,0)$ is said to be bisected by the parabola $P:y^2=8x$ if the midpoint of that chord lies on the parabola.

For a circle, the midpoint of a chord cut by a line is the foot of the perpendicular from the center to that line. Since the circle is centered at $(0,0)$ , for any line through $(\alpha,0)$ , the midpoint of the chord cut from the circle is the foot of the perpendicular from the origin onto that line.

So we need all lines through $(\alpha,0)$ such that the foot of the perpendicular from $(0,0)$ to the line lies on the parabola.

We want three distinct such lines.

Equation of a variable line through $(\alpha,0)$

Take a non-vertical line through $(\alpha,0)$ : $y=m(x-\alpha)$ which can be written as $mx-y-m\alpha=0.$

The foot of the perpendicular from the origin $(0,0)$ to the line $Ax+By+C=0$ is $\left(\frac{-AC}{A^2+B^2},\frac{-BC}{A^2+B^2}\right).$

Here, $A=m,\quad B=-1,\quad C=-m\alpha.$ Thus the foot is

=\left(\frac{m^2\alpha}{m^2+1},\,-\frac{m\alpha}{m^2+1}\right).$$ Call this point $(x_0,y_0)$. --- 3. **Condition that midpoint lies on the parabola** Since $(x_0,y_0)$ lies on $$y^2=8x,$$ we must have $$y_0^2=8x_0.$$ Substitute: $$\left(-\frac{m\alpha}{m^2+1}\right)^2=8\left(\frac{m^2\alpha}{m^2+1}\right).$$ So, $$\frac{m^2\alpha^2}{(m^2+1)^2}=\frac{8m^2\alpha}{m^2+1}.$$ Multiply by $(m^2+1)^2$: $$m^2\alpha^2=8m^2\alpha(m^2+1).$$ Hence $$m^2\alpha\big(\alpha-8(m^2+1)\big)=0.$$ --- 4. **Count the number of lines** We need distinct lines through $(\alpha,0)$ satisfying this. From $$m^2\alpha\big(\alpha-8(m^2+1)\big)=0,$$ we get possibilities: ### (i) $m=0$ This gives the horizontal line $$y=0.$$ It always works as long as the corresponding chord exists. For the circle $x^2+y^2=4$, the line $y=0$ cuts the circle in a diameter, so this is always one valid chord. ### (ii) $\alpha=0$ Then every slope seems to satisfy the equation, but all such lines pass through the origin. The midpoint of every chord through the center is the center $(0,0)$, and $(0,0)$ is on the parabola. So infinitely many lines work. But the question asks for values of $\alpha$ for which **three chords on three distinct lines** are bisected by the parabola. Thus $\alpha=0$ certainly qualifies. However, since the set is said to be an interval $(p,q)$, this suggests we should determine the open interval of nonzero values producing at least three lines, and check the boundary behavior. ### (iii) $\alpha=8(m^2+1)$ Then $$m^2=\frac{\alpha}{8}-1.$$ So real nonzero slopes exist iff $$\frac{\alpha}{8}-1>0 \quad\Longleftrightarrow\quad \alpha>8.$$ For each such $\alpha$, we get $$m=\pm\sqrt{\frac{\alpha}{8}-1},$$ which gives two distinct non-horizontal lines, plus the line $y=0$. So for $\alpha>8$, we get exactly three distinct lines. --- 5. **But these lines must cut the circle in chords** A line through $(\alpha,0)$ need not intersect the circle if $(\alpha,0)$ is far away. We must ensure each of these three lines actually meets the circle. A line through $(\alpha,0)$ with slope $m$ is $$mx-y-m\alpha=0.$$ Its distance from the origin is $$d=\frac{|m\alpha|}{\sqrt{m^2+1}}.$$ For it to cut the circle $x^2+y^2=4$ (radius $2$), we need $$d<2$$ or at least $d\le 2$ if tangent counted; but tangent gives a degenerate chord, so for a proper chord we need $d<2$. For the special slopes satisfying $$\alpha=8(m^2+1),$$ we have $$m^2=\frac{\alpha}{8}-1.$$ Then $$m^2+1=\frac{\alpha}{8}.$$ So $$d^2=\frac{m^2\alpha^2}{m^2+1}= rac{\left(\frac{\alpha}{8}-1\right)\alpha^2}{\alpha/8} =\alpha\left(\alpha-8\right).$$ Thus the chord condition is $$\alpha(\alpha-8)<4.$$ So, $$\alpha^2-8\alpha-4<0.$$ The roots are $$\alpha=4\pm 2\sqrt{5}.$$ Hence $$4-2\sqrt5<\alpha<4+2\sqrt5.$$ But from existence of the two extra slopes we also need $\alpha>8$. Therefore, $$8<\alpha<4+2\sqrt5.$$ This is impossible because $$4+2\sqrt5\approx 8.472,$$ so actually it is possible and gives a small interval. Thus for nonzero $\alpha$, exactly three distinct chord-lines occur when $$\alpha\in(8,\,4+2\sqrt5).$$ At $\alpha=0$, infinitely many lines work, but this isolated point is not part of the interval mentioned in the problem. So the interval is $$(p,q)=(8,\,4+2\sqrt5).$$ --- 6. **Compute $(2q-p)^2$** $$2q-p=2(4+2\sqrt5)-8=4\sqrt5.$$ Therefore, $$ (2q-p)^2=(4\sqrt5)^2=80.$$ --- 7. **Final answer** $$\boxed{80}$$

More from Parabola

View written solutionFree

Correct answer: 80

Interpret the condition

A chord of the circle $C:x^2+y^2=4$ lying on a line through $(\alpha,0)$ is said to be bisected by the parabola $P:y^2=8x$ if the midpoint of that chord lies on the parabola.

So we need all lines through $(\alpha,0)$ such that the foot of the perpendicular from $(0,0)$ to the line lies on the parabola.

We want three distinct such lines.

Equation of a variable line through $(\alpha,0)$

Take a non-vertical line through $(\alpha,0)$ : $y=m(x-\alpha)$ which can be written as $mx-y-m\alpha=0.$

The foot of the perpendicular from the origin $(0,0)$ to the line $Ax+By+C=0$ is $\left(\frac{-AC}{A^2+B^2},\frac{-BC}{A^2+B^2}\right).$

Here, $A=m,\quad B=-1,\quad C=-m\alpha.$ Thus the foot is

=\left(\frac{m^2\alpha}{m^2+1},\,-\frac{m\alpha}{m^2+1}\right).$$ Call this point $(x_0,y_0)$. --- 3. **Condition that midpoint lies on the parabola** Since $(x_0,y_0)$ lies on $$y^2=8x,$$ we must have $$y_0^2=8x_0.$$ Substitute: $$\left(-\frac{m\alpha}{m^2+1}\right)^2=8\left(\frac{m^2\alpha}{m^2+1}\right).$$ So, $$\frac{m^2\alpha^2}{(m^2+1)^2}=\frac{8m^2\alpha}{m^2+1}.$$ Multiply by $(m^2+1)^2$: $$m^2\alpha^2=8m^2\alpha(m^2+1).$$ Hence $$m^2\alpha\big(\alpha-8(m^2+1)\big)=0.$$ --- 4. **Count the number of lines** We need distinct lines through $(\alpha,0)$ satisfying this. From $$m^2\alpha\big(\alpha-8(m^2+1)\big)=0,$$ we get possibilities: ### (i) $m=0$ This gives the horizontal line $$y=0.$$ It always works as long as the corresponding chord exists. For the circle $x^2+y^2=4$, the line $y=0$ cuts the circle in a diameter, so this is always one valid chord. ### (ii) $\alpha=0$ Then every slope seems to satisfy the equation, but all such lines pass through the origin. The midpoint of every chord through the center is the center $(0,0)$, and $(0,0)$ is on the parabola. So infinitely many lines work. But the question asks for values of $\alpha$ for which **three chords on three distinct lines** are bisected by the parabola. Thus $\alpha=0$ certainly qualifies. However, since the set is said to be an interval $(p,q)$, this suggests we should determine the open interval of nonzero values producing at least three lines, and check the boundary behavior. ### (iii) $\alpha=8(m^2+1)$ Then $$m^2=\frac{\alpha}{8}-1.$$ So real nonzero slopes exist iff $$\frac{\alpha}{8}-1>0 \quad\Longleftrightarrow\quad \alpha>8.$$ For each such $\alpha$, we get $$m=\pm\sqrt{\frac{\alpha}{8}-1},$$ which gives two distinct non-horizontal lines, plus the line $y=0$. So for $\alpha>8$, we get exactly three distinct lines. --- 5. **But these lines must cut the circle in chords** A line through $(\alpha,0)$ need not intersect the circle if $(\alpha,0)$ is far away. We must ensure each of these three lines actually meets the circle. A line through $(\alpha,0)$ with slope $m$ is $$mx-y-m\alpha=0.$$ Its distance from the origin is $$d=\frac{|m\alpha|}{\sqrt{m^2+1}}.$$ For it to cut the circle $x^2+y^2=4$ (radius $2$), we need $$d<2$$ or at least $d\le 2$ if tangent counted; but tangent gives a degenerate chord, so for a proper chord we need $d<2$. For the special slopes satisfying $$\alpha=8(m^2+1),$$ we have $$m^2=\frac{\alpha}{8}-1.$$ Then $$m^2+1=\frac{\alpha}{8}.$$ So $$d^2=\frac{m^2\alpha^2}{m^2+1}= rac{\left(\frac{\alpha}{8}-1\right)\alpha^2}{\alpha/8} =\alpha\left(\alpha-8\right).$$ Thus the chord condition is $$\alpha(\alpha-8)<4.$$ So, $$\alpha^2-8\alpha-4<0.$$ The roots are $$\alpha=4\pm 2\sqrt{5}.$$ Hence $$4-2\sqrt5<\alpha<4+2\sqrt5.$$ But from existence of the two extra slopes we also need $\alpha>8$. Therefore, $$8<\alpha<4+2\sqrt5.$$ This is impossible because $$4+2\sqrt5\approx 8.472,$$ so actually it is possible and gives a small interval. Thus for nonzero $\alpha$, exactly three distinct chord-lines occur when $$\alpha\in(8,\,4+2\sqrt5).$$ At $\alpha=0$, infinitely many lines work, but this isolated point is not part of the interval mentioned in the problem. So the interval is $$(p,q)=(8,\,4+2\sqrt5).$$ --- 6. **Compute $(2q-p)^2$** $$2q-p=2(4+2\sqrt5)-8=4\sqrt5.$$ Therefore, $$ (2q-p)^2=(4\sqrt5)^2=80.$$ --- 7. **Final answer** $$\boxed{80}$$

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