Question

From the point P(h, k) three normals are drawn to the parabola $x^2 = 8y$ and $m_1, m_2$ and $m_3$ are the slopes of three normals

Answer 1

a) 0, b) 2, c) y = 3/2

Answer 2

The parabola is given by $x^2 = 8y$. This is of the form $x^2 = 4ay$, where $4a = 8$, so $a = 2$.

The equation of a normal to the parabola $x^2 = 4ay$ with slope $m$ is given by: $x = my - 2am - am^3$. Substituting $a=2$, we get: $x = my - 2(2)m - 2m^3$ $x = my - 4m - 2m^3$.

Since the normal passes through the point P(h, k), we substitute (h, k) into the equation: $h = mk - 4m - 2m^3$. Rearranging this into a cubic equation in $m$: $2m^3 + (4-k)m + h = 0$. Let $m_1, m_2, m_3$ be the slopes of the three normals drawn from P(h, k). These are the roots of the cubic equation.

For a cubic equation $Am^3 + Bm^2 + Cm + D = 0$: Sum of roots: $m_1 + m_2 + m_3 = -B/A$ Sum of product of roots taken two at a time: $m_1m_2 + m_2m_3 + m_3m_1 = C/A$ Product of roots: $m_1m_2m_3 = -D/A$

In our equation $2m^3 + 0m^2 + (4-k)m + h = 0$: $A=2$, $B=0$, $C=(4-k)$, $D=h$.

(a) Find the algebraic sum of the slopes of these three normals. The algebraic sum of the slopes is $m_1 + m_2 + m_3 = -B/A = -0/2 = 0$.

(b) If two of the three normals are at right angles then the locus of point P is a conic, find the latus rectum of conic. If two of the normals are at right angles, let $m_1m_2 = -1$. From the cubic equation, the product of the roots is $m_1m_2m_3 = -D/A = -h/2$. Substitute $m_1m_2 = -1$ into the product of roots equation: $(-1)m_3 = -h/2$ $m_3 = h/2$.

Since $m_3$ is a root of the cubic equation, it must satisfy $2m^3 + (4-k)m + h = 0$. Substitute $m = h/2$: $2(h/2)^3 + (4-k)(h/2) + h = 0$ $2(h^3/8) + (4-k)h/2 + h = 0$ $h^3/4 + (4-k)h/2 + h = 0$.

Factor out $h$: $h(h^2/4 + (4-k)/2 + 1) = 0$. This implies $h=0$ or $h^2/4 + (4-k)/2 + 1 = 0$.

If $h=0$, then $m_3=0$. From $m_1+m_2+m_3=0$, we get $m_1+m_2=0$, so $m_2=-m_1$. Given $m_1m_2=-1$, we have $m_1(-m_1)=-1$, so $m_1^2=1$, which means $m_1 = \pm 1$. The cubic equation becomes $2m^3 + (4-k)m = 0$, or $m(2m^2 + 4-k) = 0$. The roots are $m=0$ and $m^2 = (k-4)/2$. For $m^2=1$, we need $(k-4)/2 = 1$, so $k-4=2$, which gives $k=6$. So $(0,6)$ is a point on the locus.

Now consider the other case: $h^2/4 + (4-k)/2 + 1 = 0$. Multiply by 4 to clear the denominators: $h^2 + 2(4-k) + 4 = 0$ $h^2 + 8 - 2k + 4 = 0$ $h^2 - 2k + 12 = 0$ $h^2 = 2k - 12$ $h^2 = 2(k - 6)$.

This is the locus of P(h, k). Replacing $(h, k)$ with $(x, y)$, the locus is: $x^2 = 2(y - 6)$. This is a parabola of the form $X^2 = 4AY$, where $X=x$, $Y=y-6$, and $4A=2$. Thus, $A = 1/2$. The latus rectum of a parabola $X^2 = 4AY$ is $4A$. So, the latus rectum of the conic is $2$.

(c) If the two normals from P are such that they make complementary angles with the axis then the locus of point P is a conic, find a directrix of conic. The axis of the parabola $x^2 = 8y$ is the y-axis. If two lines make complementary angles with the y-axis, let the angles be $\theta_1$ and $\theta_2$, such that $\theta_1 + \theta_2 = \pi/2$. The slopes of these normals are $m_1 = \tan(\pi/2 - \theta_1) = \cot\theta_1$ and $m_2 = \tan(\pi/2 - \theta_2) = \cot\theta_2$. Since $\theta_2 = \pi/2 - \theta_1$, we have $\cot\theta_2 = \cot(\pi/2 - \theta_1) = \tan\theta_1$. Therefore, $m_1 = \cot\theta_1$ and $m_2 = \tan\theta_1$. The product of these slopes is $m_1m_2 = (\cot\theta_1)(\tan\theta_1) = 1$.

From the cubic equation, we have $m_1m_2m_3 = -h/2$. Substitute $m_1m_2 = 1$: $(1)m_3 = -h/2$ $m_3 = -h/2$.

Since $m_3$ is a root of the cubic equation, it must satisfy $2m^3 + (4-k)m + h = 0$. Substitute $m = -h/2$: $2(-h/2)^3 + (4-k)(-h/2) + h = 0$ $2(-h^3/8) - (4-k)h/2 + h = 0$ $-h^3/4 - (4-k)h/2 + h = 0$.

Factor out $h$: $h(-h^2/4 - (4-k)/2 + 1) = 0$. This implies $h=0$ or $-h^2/4 - (4-k)/2 + 1 = 0$.

If $h=0$, then $m_3=0$. From $m_1+m_2+m_3=0$, we get $m_1+m_2=0$, so $m_2=-m_1$. Given $m_1m_2=1$, we have $m_1(-m_1)=1$, so $-m_1^2=1$, which means $m_1^2=-1$. This leads to imaginary slopes, which are not possible for real normals. Thus, $h \neq 0$.

So, we must have $-h^2/4 - (4-k)/2 + 1 = 0$. Multiply by -4 to clear the denominators and make $h^2$ positive: $h^2 + 2(4-k) - 4 = 0$ $h^2 + 8 - 2k - 4 = 0$ $h^2 - 2k + 4 = 0$ $h^2 = 2k - 4$ $h^2 = 2(k - 2)$.

This is the locus of P(h, k). Replacing $(h, k)$ with $(x, y)$, the locus is: $x^2 = 2(y - 2)$. This is a parabola of the form $X^2 = 4AY$, where $X=x$, $Y=y-2$, and $4A=2$. Thus, $A = 1/2$. The vertex of this parabola is $(0, 2)$. Since $x^2 = 4AY$ opens upwards, the directrix is $Y = Y_0 - A$. Here $Y_0=2$ and $A=1/2$. So, the directrix is $y = 2 - 1/2 = 3/2$.

The final answer is $\boxed{a) 0, b) 2, c) y = 3/2}$

Explanation of the solution:

1. Normal Equation: The general normal equation for $x^2=4ay$ is $x=my-2am-am^3$. For $x^2=8y$, $a=2$, so $x=my-4m-2m^3$.
2. Cubic Equation for Slopes: Since the normal passes through P(h,k), substitute (h,k) into the normal equation to get the cubic equation for slopes: $2m^3 + (4-k)m + h = 0$.
3. Part (a) - Sum of Slopes: For $Am^3+Bm^2+Cm+D=0$, the sum of roots is $-B/A$. Here, $B=0$, so $m_1+m_2+m_3=0$.
4. Part (b) - Right Angles: If $m_1m_2=-1$. The product of roots is $m_1m_2m_3 = -D/A = -h/2$. So, $(-1)m_3 = -h/2 \implies m_3 = h/2$. Substitute $m_3$ into the cubic equation and simplify to get the locus $x^2=2(y-6)$. This is a parabola $X^2=4AY$ with $4A=2$, so its latus rectum is $2$.
5. Part (c) - Complementary Angles: "Complementary angles with the axis" (y-axis for $x^2=8y$) means if angles are $\theta_1, \theta_2$ with y-axis, then $\theta_1+\theta_2=\pi/2$. Slopes are $m_1=\cot\theta_1$ and $m_2=\cot\theta_2=\tan\theta_1$. So $m_1m_2=1$. The product of roots is $m_1m_2m_3 = -h/2$. So, $(1)m_3 = -h/2 \implies m_3 = -h/2$. Substitute $m_3$ into the cubic equation and simplify to get the locus $x^2=2(y-2)$. This is a parabola $X^2=4AY$ with $4A=2$, so $A=1/2$. Its vertex is $(0,2)$. The directrix for $x^2=4A(y-Y_0)$ is $y=Y_0-A$, so $y=2-1/2=3/2$.

Answer:

(a) The algebraic sum of the slopes of these three normals is $\mathbf{0}$. (b) The latus rectum of the conic is $\mathbf{2}$. (c) A directrix of the conic is $\mathbf{y = 3/2}$.

Subject, Chapter and Topic:

Mathematics, Conic Sections, Parabola, Equation of Normal of a Parabola.

Difficulty Level:

Medium

Question Type:

Descriptive