Question

If the locus of the middle points of all such chords of the parabola $y^2=8x$ which has a length of 2 units, has the equation $y^4+ly^2(2-x)=mx-n$ where $l,m,n \in N$, find the value of $(l+m+n)$.

Answer 1

152

Answer 2

To find the locus of the middle points of chords of the parabola $y^2=8x$ with a length of 2 units, we can use the following approach:

Let the midpoint of a chord be $P(h,k)$. For a parabola $y^2=4ax$, the equation of the chord with midpoint $(h,k)$ is given by $T=S_1$, where $T$ is the tangent at $(h,k)$ and $S_1$ is the value of the parabola equation at $(h,k)$. Given parabola is $y^2=8x$. Comparing with $y^2=4ax$, we have $4a=8 \implies a=2$.

The equation of the chord is: $yk - 4(x+h) = k^2 - 8h$ $yk - 4x - 4h = k^2 - 8h$ $yk = 4x + k^2 - 4h$

To find the points of intersection of this chord with the parabola, substitute $x = \frac{yk - k^2 + 4h}{4}$ into $y^2=8x$: $y^2 = 8 \left( \frac{yk - k^2 + 4h}{4} \right)$ $y^2 = 2(yk - k^2 + 4h)$ $y^2 - 2ky + (2k^2 - 8h) = 0$

Let the roots of this quadratic equation in $y$ be $y_1$ and $y_2$. These are the y-coordinates of the endpoints of the chord. From Vieta's formulas: $y_1+y_2 = 2k$ $y_1y_2 = 2k^2 - 8h$

The x-coordinates of the endpoints are $x_1 = y_1^2/8$ and $x_2 = y_2^2/8$.

The length of the chord $L$ is given as 2 units. So $L^2=4$. $L^2 = (x_2-x_1)^2 + (y_2-y_1)^2$

First, find $(y_2-y_1)^2$: $(y_2-y_1)^2 = (y_1+y_2)^2 - 4y_1y_2$ $(y_2-y_1)^2 = (2k)^2 - 4(2k^2 - 8h)$ $(y_2-y_1)^2 = 4k^2 - 8k^2 + 32h$ $(y_2-y_1)^2 = 32h - 4k^2$

Next, find $(x_2-x_1)^2$: $x_2-x_1 = \frac{y_2^2}{8} - \frac{y_1^2}{8} = \frac{1}{8}(y_2^2-y_1^2) = \frac{1}{8}(y_2-y_1)(y_2+y_1)$ Substitute $y_1+y_2 = 2k$: $x_2-x_1 = \frac{1}{8}(y_2-y_1)(2k) = \frac{k}{4}(y_2-y_1)$ $(x_2-x_1)^2 = \frac{k^2}{16}(y_2-y_1)^2$

Now substitute these expressions into the length formula $L^2=4$: $4 = \frac{k^2}{16}(y_2-y_1)^2 + (y_2-y_1)^2$ $4 = (y_2-y_1)^2 \left( \frac{k^2}{16} + 1 \right)$ Substitute $(y_2-y_1)^2 = 32h - 4k^2$: $4 = (32h - 4k^2) \left( \frac{k^2+16}{16} \right)$ Factor out 4 from the first term: $4 = 4(8h - k^2) \left( \frac{k^2+16}{16} \right)$ Divide both sides by 4: $1 = (8h - k^2) \left( \frac{k^2+16}{16} \right)$ Multiply both sides by 16: $16 = (8h - k^2)(k^2+16)$

This is the locus of the midpoint $(h,k)$. Replace $h$ with $x$ and $k$ with $y$: $16 = (8x - y^2)(y^2+16)$ Expand the right side: $16 = 8xy^2 + 128x - y^4 - 16y^2$

Rearrange the terms to match the given equation $y^4+ly^2(2-x)=mx-n$: $y^4 + 16y^2 - 8xy^2 - 128x + 16 = 0$ $y^4 + y^2(16 - 8x) = 128x - 16$ $y^4 + 8y^2(2 - x) = 128x - 16$

Comparing this with $y^4+ly^2(2-x)=mx-n$: $l=8$ $m=128$ $n=16$

The problem states $l,m,n \in N$ (natural numbers). All values $l=8, m=128, n=16$ are natural numbers.

Finally, calculate the value of $(l+m+n)$: $l+m+n = 8 + 128 + 16 = 152$.