Question

Locus of the centre of the circle passing through the vertex and the mid-points of perpendicular chords from the vertex of the parabola $y^2=4ax$ is

• A. is a parabola with vertex (-a, a)
• B. is a parabola with latus rectum a
• C. is a parabola with vertex (2a, 0)
• D. is a parabola with latus rectum $\frac{a}{2}$

Answer 1

Answer: (B), (C)

is a parabola with latus rectum a and is a parabola with vertex (2a, 0)

Answer 2

Let the equation of the parabola be $y^2 = 4ax$. Its vertex is $V(0,0)$.
Let the equation of the circle be $x^2 + y^2 - 2hx - 2ky + c = 0$.
Since the circle passes through the vertex $V(0,0)$, substituting $(0,0)$ into the circle's equation gives:
$0^2 + 0^2 - 2h(0) - 2k(0) + c = 0 \implies c = 0$.
So, the equation of the circle is $x^2 + y^2 - 2hx - 2ky = 0$. The center of this circle is $(h,k)$.

Consider two chords from the vertex, $VP$ and $VQ$, which are perpendicular.
Let $P = (at_1^2, 2at_1)$ and $Q = (at_2^2, 2at_2)$ be points on the parabola.
The slope of $VP$ is $m_1 = \frac{2at_1 - 0}{at_1^2 - 0} = \frac{2}{t_1}$.
The slope of $VQ$ is $m_2 = \frac{2at_2 - 0}{at_2^2 - 0} = \frac{2}{t_2}$.
Since $VP \perp VQ$, their slopes satisfy $m_1 m_2 = -1$:
$\left(\frac{2}{t_1}\right) \left(\frac{2}{t_2}\right) = -1 \implies \frac{4}{t_1 t_2} = -1 \implies t_1 t_2 = -4$.

The circle passes through the mid-points of these perpendicular chords.
Let $M_1$ be the midpoint of $VP$: $M_1 = \left(\frac{0+at_1^2}{2}, \frac{0+2at_1}{2}\right) = \left(\frac{at_1^2}{2}, at_1\right)$.
Let $M_2$ be the midpoint of $VQ$: $M_2 = \left(\frac{0+at_2^2}{2}, \frac{0+2at_2}{2}\right) = \left(\frac{at_2^2}{2}, at_2\right)$.

Since $M_1$ lies on the circle $x^2 + y^2 - 2hx - 2ky = 0$:
$\left(\frac{at_1^2}{2}\right)^2 + (at_1)^2 - 2h\left(\frac{at_1^2}{2}\right) - 2k(at_1) = 0$
$\frac{a^2t_1^4}{4} + a^2t_1^2 - hat_1^2 - 2kat_1 = 0$.
Assuming $t_1 \neq 0$ (if $t_1=0$, $P$ is the vertex, $VP$ is not a chord in the sense of a line segment for which we can find a midpoint distinct from V), we can divide by $at_1$:
$\frac{at_1^3}{4} + at_1 - ht_1 - 2k = 0$
$\frac{a}{4}t_1^3 + (a-h)t_1 = 2k$ (Equation 1)

Similarly, for $M_2$ lying on the circle:
$\frac{a}{4}t_2^3 + (a-h)t_2 = 2k$ (Equation 2)

Subtract Equation 2 from Equation 1:
$\frac{a}{4}(t_1^3 - t_2^3) + (a-h)(t_1 - t_2) = 0$.
Since $t_1 \neq t_2$ (otherwise $P=Q$, which means $VP$ and $VQ$ are the same chord, and cannot be perpendicular unless $a=0$), we can divide by $(t_1 - t_2)$:
$\frac{a}{4}(t_1^2 + t_1 t_2 + t_2^2) + (a-h) = 0$.
Substitute $t_1 t_2 = -4$:
$\frac{a}{4}(t_1^2 - 4 + t_2^2) + (a-h) = 0$
$\frac{a}{4}((t_1+t_2)^2 - 2t_1t_2 - 4) + (a-h) = 0$
$\frac{a}{4}((t_1+t_2)^2 - 2(-4) - 4) + (a-h) = 0$
$\frac{a}{4}((t_1+t_2)^2 + 8 - 4) + (a-h) = 0$
$\frac{a}{4}((t_1+t_2)^2 + 4) + (a-h) = 0$.
Multiply by 4:
$a((t_1+t_2)^2 + 4) + 4(a-h) = 0$
$a(t_1+t_2)^2 + 4a + 4a - 4h = 0$
$a(t_1+t_2)^2 = 4h - 8a$ (Equation 3)

Now, multiply Equation 1 by $t_2$:
$\frac{a}{4}t_1^3 t_2 + (a-h)t_1 t_2 = 2kt_2$.
Substitute $t_1 t_2 = -4$:
$\frac{a}{4}t_1^2(-4) + (a-h)(-4) = 2kt_2$
$-at_1^2 - 4(a-h) = 2kt_2$ (Equation 4)

Multiply Equation 2 by $t_1$:
$\frac{a}{4}t_2^3 t_1 + (a-h)t_2 t_1 = 2kt_1$.
Substitute $t_1 t_2 = -4$:
$\frac{a}{4}t_2^2(-4) + (a-h)(-4) = 2kt_1$
$-at_2^2 - 4(a-h) = 2kt_1$ (Equation 5)

Add Equation 4 and Equation 5:
$-a(t_1^2 + t_2^2) - 8(a-h) = 2k(t_1+t_2)$
$-a((t_1+t_2)^2 - 2t_1t_2) - 8(a-h) = 2k(t_1+t_2)$
$-a((t_1+t_2)^2 - 2(-4)) - 8(a-h) = 2k(t_1+t_2)$
$-a((t_1+t_2)^2 + 8) - 8(a-h) = 2k(t_1+t_2)$ (Equation 6)

Let $S = t_1+t_2$.
From Equation 3: $aS^2 = 4h - 8a$.
From Equation 6: $-a(S^2+8) - 8(a-h) = 2kS$.
Substitute $S^2 = \frac{4h-8a}{a}$ into Equation 6:
$-a\left(\frac{4h-8a}{a} + 8\right) - 8a + 8h = 2kS$
$-(4h-8a + 8a) - 8a + 8h = 2kS$
$-4h - 8a + 8h = 2kS$
$4h - 8a = 2kS$
$2h - 4a = kS$.

If $k \neq 0$, then $S = \frac{2h-4a}{k}$.
Substitute this expression for $S$ into $aS^2 = 4h-8a$:
$a\left(\frac{2h-4a}{k}\right)^2 = 4h-8a$
$a\frac{4(h-2a)^2}{k^2} = 4(h-2a)$.
If $h-2a \neq 0$, we can divide by $4(h-2a)$:
$a\frac{h-2a}{k^2} = 1 \implies k^2 = a(h-2a)$.

If $h-2a = 0$, then $h=2a$. From $aS^2 = 4h-8a$, we get $aS^2 = 4(2a)-8a=0$, so $S=0$.
From $2h-4a=kS$, we get $2(2a)-4a=k(0)$, which is $0=0$.
From Equation 1: $\frac{a}{4}t_1^3 + (a-h)t_1 = 2k$. Substitute $h=2a$:
$\frac{a}{4}t_1^3 + (a-2a)t_1 = 2k \implies \frac{a}{4}t_1^3 - at_1 = 2k$.
Since $S=t_1+t_2=0$, $t_2=-t_1$. And $t_1 t_2 = -4 \implies t_1(-t_1)=-4 \implies t_1^2=4$.
So $t_1^3 = t_1 \cdot t_1^2 = 4t_1$.
Substituting $t_1^3=4t_1$ into the equation for $k$:
$\frac{a}{4}(4t_1) - at_1 = 2k \implies at_1 - at_1 = 2k \implies 0 = 2k \implies k=0$.
So, when $h=2a$, $k=0$. The point $(2a,0)$ satisfies $k^2 = a(h-2a)$, as $0^2 = a(2a-2a)=0$.
Thus, the locus of the center $(h,k)$ is $k^2 = a(h-2a)$.
Replacing $(h,k)$ with $(x,y)$, the locus is $y^2 = a(x-2a)$.

This is the equation of a parabola.
Its vertex is $(2a,0)$.
Its latus rectum is $a$.

The locus of the centre of the circle is the parabola $y^2 = a(x-2a)$.
This parabola has:

• Vertex at $(2a,0)$.
• Latus rectum equal to $a$. Therefore, options (B) and (C) are correct.