Question

A variable chord of the parabola $y^2=8x$ touches the parabola $y^2=2x$. The locus of the point of intersection of the tangent at the end of the chord is a parabola. Find its latus rectum.

• A. 8
• B. 16
• C. 32
• D. 64

Answer 1

Answer: (C)

32

Answer 2

Let the first parabola be $P_1: y^2 = 8x$. This is of the form $y^2 = 4a_1x$, so $4a_1 = 8$, which gives $a_1 = 2$. Let the second parabola be $P_2: y^2 = 2x$. This is of the form $y^2 = 4a_2x$, so $4a_2 = 2$, which gives $a_2 = 1/2$.

Let $T(h, k)$ be a point on the locus. $T(h,k)$ is the point of intersection of the tangents at the extremities of a chord of $P_1$. The equation of the chord of contact of tangents from an external point $(h, k)$ to the parabola $y^2 = 4a_1x$ is given by $yk = 2a_1(x+h)$. For $P_1: y^2 = 8x$ ($a_1=2$), the chord of contact from $T(h,k)$ is: $yk = 2(2)(x+h)$ $yk = 4(x+h)$

This chord $yk = 4x + 4h$ touches the parabola $P_2: y^2 = 2x$. We can rewrite the chord equation as $y = \frac{4}{k}x + \frac{4h}{k}$. This is a line of the form $y = mx + c$, where $m = \frac{4}{k}$ and $c = \frac{4h}{k}$.

The condition for a line $y = mx+c$ to touch a parabola $y^2 = 4ax$ is $c = \frac{a}{m^2}$. For parabola $P_2: y^2 = 2x$, we have $a_2 = 1/2$. So, the condition for the chord to touch $P_2$ is: $c = \frac{a_2}{m^2}$ $\frac{4h}{k} = \frac{1/2}{(4/k)^2}$

Let's evaluate the right side: $\frac{1/2}{(4/k)^2} = \frac{1/2}{16/k^2} = \frac{1}{2} \cdot \frac{k^2}{16} = \frac{k^2}{32}$

Now, substitute this back into the condition: $\frac{4h}{k} = \frac{k^2}{32}$

We must consider the case where $k=0$. If $k=0$, the point is $T(h,0)$. The chord of contact equation becomes $0 \cdot y = 4(x+h)$, which simplifies to $x = -h$. This is a vertical line. A vertical line $x=X$ is tangent to $y^2=2x$ only if $X=0$ (at the vertex $(0,0)$). Thus, if $k=0$, then $-h=0$, which implies $h=0$. So, the point $(0,0)$ is a possible point on the locus.

If $k \neq 0$, we can proceed with the equation $\frac{4h}{k} = \frac{k^2}{32}$. Multiply both sides by $32k$: $4h \cdot 32 = k \cdot k^2$ $128h = k^3$

This equation $k^3 = 128h$ relates the coordinates $(h,k)$ of the point $T$. The locus is $y^3 = 128x$. This is a cubic curve, not a parabola, which contradicts the problem statement that the locus is a parabola.

Let's re-evaluate the tangent condition using the parametric form of the tangent. The tangent to $y^2=2x$ ($a_2=1/2$) at parameter $t$ is $y = tx + \frac{a_2}{t}$, which is $y = tx + \frac{1}{2t}$. This line $y = tx + \frac{1}{2t}$ must be identical to the chord $y = \frac{4}{k}x + \frac{4h}{k}$.

Comparing the coefficients of $x$ and the constant terms:

1. $m = t = \frac{4}{k}$
2. $c = \frac{1}{2t} = \frac{4h}{k}$

Substitute the value of $t$ from (1) into (2): $\frac{1}{2(4/k)} = \frac{4h}{k}$ $\frac{k}{8} = \frac{4h}{k}$

Now, we solve for $h$ and $k$. Assuming $k \neq 0$ (as seen before, $k=0$ implies $h=0$, which is the origin): $k \cdot k = 8 \cdot 4h$ $k^2 = 32h$

This equation $k^2 = 32h$ represents the locus of the point $T(h,k)$. The locus is the parabola $y^2 = 32x$. This is in the standard form $y^2 = 4ax$, where $4a = 32$. The latus rectum of a parabola $y^2 = 4ax$ is $4a$.

Therefore, the latus rectum of the locus parabola $y^2 = 32x$ is 32.

The case $k=0$ leads to $h=0$, giving the point $(0,0)$. This point satisfies $k^2=32h$ as $0^2 = 32(0)$. So the locus equation holds for the origin as well.