Question

A KCLS parabola touches straight lines x - 2y +13 = 0 and 2x+y+1= 0 at points P(17, 15) and Q(2, -5) respectively then:

Answer 1

Focus is (-1/2, 25/2) or (39/2, -5/2)

Answer 2

The problem states that a parabola touches two straight lines $L_1: x - 2y + 13 = 0$ and $L_2: 2x + y + 1 = 0$ at points $P(17, 15)$ and $Q(2, -5)$ respectively.

1. Find the intersection point of the tangents: Let $R$ be the intersection point of $L_1$ and $L_2$. $x - 2y + 13 = 0 \quad (1)$ $2x + y + 1 = 0 \quad (2)$ From (2), $y = -2x - 1$. Substitute into (1): $x - 2(-2x - 1) + 13 = 0 \implies x + 4x + 2 + 13 = 0 \implies 5x + 15 = 0 \implies x = -3$. Substitute $x = -3$ into (2): $2(-3) + y + 1 = 0 \implies -6 + y + 1 = 0 \implies y = 5$. So, $R(-3, 5)$.

2. Determine the relationship between the tangents: The slope of $L_1$ is $m_1 = 1/2$. The slope of $L_2$ is $m_2 = -2$. Since $m_1 m_2 = (1/2)(-2) = -1$, the two tangents $L_1$ and $L_2$ are perpendicular.

3. Use properties of perpendicular tangents: If two tangents to a parabola are perpendicular, their intersection point $R$ lies on the directrix of the parabola. Also, the line segment connecting the focus $S$ to the intersection point of perpendicular tangents $R$ is the axis of the parabola. This means $SR$ is the axis of the parabola. Since $R$ is on the directrix and $SR$ is the axis, the directrix must be perpendicular to $SR$ and pass through $R$.

4. Find the equations of the angle bisectors of the tangents: The line $SR$ (the axis) is one of the angle bisectors of the tangents $L_1$ and $L_2$. The directrix is the other angle bisector. The equations of the angle bisectors are given by: $\frac{x - 2y + 13}{\sqrt{1^2 + (-2)^2}} = \pm \frac{2x + y + 1}{\sqrt{2^2 + 1^2}}$ $\frac{x - 2y + 13}{\sqrt{5}} = \pm \frac{2x + y + 1}{\sqrt{5}}$ $x - 2y + 13 = \pm (2x + y + 1)$

   Case 1 (Angle Bisector 1): $x - 2y + 13 = 2x + y + 1 \implies -x - 3y + 12 = 0 \implies x + 3y - 12 = 0$. Case 2 (Angle Bisector 2): $x - 2y + 13 = -(2x + y + 1) \implies x - 2y + 13 = -2x - y - 1 \implies 3x - y + 14 = 0$. Both these lines pass through $R(-3, 5)$.

5. Use the property of the circumcircle of $\triangle RPQ$: The circumcircle of the triangle formed by two tangents and their chord of contact passes through the focus. Thus, the circumcircle of $\triangle RPQ$ passes through the focus $S(h, k)$. Since $\angle PRQ = 90^\circ$, $PQ$ is the diameter of this circumcircle. The midpoint of $PQ$ is $M = \left( \frac{17+2}{2}, \frac{15-5}{2} \right) = \left( \frac{19}{2}, 5 \right)$. The radius of the circumcircle is $r = \frac{1}{2} PQ = \frac{1}{2} \sqrt{(17-2)^2 + (15-(-5))^2} = \frac{1}{2} \sqrt{15^2 + 20^2} = \frac{1}{2} \sqrt{225 + 400} = \frac{1}{2} \sqrt{625} = \frac{25}{2}$. The equation of the circumcircle is $\left(x - \frac{19}{2}\right)^2 + (y - 5)^2 = \left(\frac{25}{2}\right)^2$. Since $S(h, k)$ lies on this circle: $\left(h - \frac{19}{2}\right)^2 + (k - 5)^2 = \frac{625}{4} \quad (3)$

6. Determine the focus and directrix: We have two possibilities for the axis and directrix:

   Possibility A: Axis is $3x - y + 14 = 0$ and Directrix is $x + 3y - 12 = 0$. The focus $S(h, k)$ lies on the axis: $3h - k + 14 = 0 \implies k = 3h + 14$. Substitute $k$ into equation (3): $\left(h - \frac{19}{2}\right)^2 + (3h + 9)^2 = \frac{625}{4}$ Multiply by 4: $(2h - 19)^2 + 4(3h + 9)^2 = 625$ $4h^2 - 76h + 361 + 4(9h^2 + 54h + 81) = 625$ $4h^2 - 76h + 361 + 36h^2 + 216h + 324 = 625$ $40h^2 + 140h + 685 = 625$ $40h^2 + 140h + 60 = 0$ Divide by 20: $2h^2 + 7h + 3 = 0$ $(2h + 1)(h + 3) = 0$ So, $h = -1/2$ or $h = -3$. If $h = -3$, then $k = 3(-3) + 14 = 5$. This gives $S(-3, 5)$, which is point $R$. A parabola cannot have its focus on its directrix unless it degenerates into a point, which is not a parabola. So $S \neq R$. Therefore, $h = -1/2$. Then $k = 3(-1/2) + 14 = -3/2 + 28/2 = 25/2$. So, the focus is $S(-1/2, 25/2)$. The directrix is $x + 3y - 12 = 0$.

   Possibility B: Axis is $x + 3y - 12 = 0$ and Directrix is $3x - y + 14 = 0$. The focus $S(h, k)$ lies on the axis: $h + 3k - 12 = 0 \implies h = 12 - 3k$. Substitute $h$ into equation (3): $\left(12 - 3k - \frac{19}{2}\right)^2 + (k - 5)^2 = \frac{625}{4}$ $\left(\frac{24 - 6k - 19}{2}\right)^2 + (k - 5)^2 = \frac{625}{4}$ $\left(\frac{5 - 6k}{2}\right)^2 + (k - 5)^2 = \frac{625}{4}$ Multiply by 4: $(5 - 6k)^2 + 4(k - 5)^2 = 625$ $25 - 60k + 36k^2 + 4(k^2 - 10k + 25) = 625$ $25 - 60k + 36k^2 + 4k^2 - 40k + 100 = 625$ $40k^2 - 100k + 125 = 625$ $40k^2 - 100k - 500 = 0$ Divide by 20: $2k^2 - 5k - 25 = 0$ Using the quadratic formula: $k = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(-25)}}{2(2)} = \frac{5 \pm \sqrt{25 + 200}}{4} = \frac{5 \pm \sqrt{225}}{4} = \frac{5 \pm 15}{4}$. So, $k = \frac{5+15}{4} = 5$ or $k = \frac{5-15}{4} = -5/2$. If $k = 5$, then $h = 12 - 3(5) = -3$. This gives $S(-3, 5)$, which is point $R$. Not possible for a non-degenerate parabola. Therefore, $k = -5/2$. Then $h = 12 - 3(-5/2) = 12 + 15/2 = 39/2$. So, the focus is $S(39/2, -5/2)$. The directrix is $3x - y + 14 = 0$.

Both solutions for the focus and directrix are valid. The question implies there is a unique parabola. However, the conditions given lead to two possible parabolas. The problem statement "A KCLS parabola" suggests a specific parabola, but mathematically, two parabolas satisfy the given conditions. Without further constraints, both are correct.

Let's assume the question expects properties of a parabola satisfying the conditions. The focus is either $(-1/2, 25/2)$ or $(39/2, -5/2)$. The directrix is either $x+3y-12=0$ or $3x-y+14=0$.

The question asks "then:". It is likely a multiple choice question where one of these properties would be an option. For example, an option might be "The focus is $(-1/2, 25/2)$" or "The directrix is $x+3y-12=0$". If the question is implicitly asking for a specific parabola, usually more information is given. However, based on the calculation, two parabolas fit the description.

Final Answer will depend on the options provided. If it's a descriptive question, both sets of focus/directrix are valid answers.

Let's verify one solution set. Parabola 1: Focus $S_1(-1/2, 25/2)$, Directrix $D_1: x+3y-12=0$. Equation of parabola: $(x - h)^2 + (y - k)^2 = \left(\frac{Ax+By+C}{\sqrt{A^2+B^2}}\right)^2$ $\left(x + \frac{1}{2}\right)^2 + \left(y - \frac{25}{2}\right)^2 = \frac{(x+3y-12)^2}{1^2+3^2}$ $\frac{(2x+1)^2}{4} + \frac{(2y-25)^2}{4} = \frac{(x+3y-12)^2}{10}$ $10((2x+1)^2 + (2y-25)^2) = 4(x+3y-12)^2$ $10(4x^2+4x+1 + 4y^2-100y+625) = 4(x^2+9y^2+144+6xy-24x-72y)$ $10(4x^2+4y^2+4x-100y+626) = 4x^2+36y^2+576+24xy-96x-288y$ $40x^2+40y^2+40x-1000y+6260 = 4x^2+36y^2+24xy-96x-288y+576$ $36x^2+4y^2-24xy+136x-712y+5684=0$ This is the equation of the parabola.

Since the question expects a specific answer, let's assume it's a multiple choice question and one of the derived properties will be an option.

The properties derived are:

1. The intersection of the tangents is $R(-3, 5)$.
2. The tangents are perpendicular.
3. $R(-3, 5)$ lies on the directrix.
4. The axis of the parabola is $SR$.
5. The circumcircle of $\triangle RPQ$ passes through the focus $S$.
6. The focus is $S_1(-1/2, 25/2)$ and the directrix is $D_1: x+3y-12=0$.
7. The focus is $S_2(39/2, -5/2)$ and the directrix is $D_2: 3x-y+14=0$.

If the question asks for the focus, there are two possible answers. If it asks for the directrix, there are two possible answers. If it asks for the axis, there are two possible answers. If it asks for the intersection of tangents, there is one answer. If it asks for the angle between tangents, there is one answer.

Let's assume the question is asking for the focus. The two possible foci are $(-1/2, 25/2)$ and $(39/2, -5/2)$.

The final answer is Focus is $\left(-\frac{1}{2}, \frac{25}{2}\right)$ or $\left(\frac{39}{2}, -\frac{5}{2}\right)$.

Explanation of the Solution:

1. Find the intersection point $R$ of the two given tangent lines $x - 2y + 13 = 0$ and $2x + y + 1 = 0$. This point is $R(-3, 5)$.
2. Calculate the slopes of the tangent lines. Slope of $L_1$ is $1/2$ and slope of $L_2$ is $-2$. Since their product is $-1$, the tangents are perpendicular.
3. A key property of parabolas states that the intersection point of two perpendicular tangents lies on the directrix. Thus, $R(-3, 5)$ is on the directrix.
4. Another property states that the line joining the focus $S$ to the intersection point of perpendicular tangents $R$ forms the axis of the parabola. Therefore, $SR$ is the axis.
5. Since the axis ($SR$) is perpendicular to the directrix, and $R$ is on the directrix, the directrix must be the line through $R$ perpendicular to $SR$. This means the axis and directrix are the two angle bisectors of the tangents $L_1$ and $L_2$. The angle bisectors are $x + 3y - 12 = 0$ and $3x - y + 14 = 0$.
6. The circumcircle of the triangle formed by the two tangents ($L_1, L_2$) and their chord of contact ($PQ$) passes through the focus $S$. Since the tangents are perpendicular, $\angle PRQ = 90^\circ$, which means $PQ$ is the diameter of this circumcircle.
7. Find the midpoint $M$ of $PQ$ as the center of the circumcircle: $M\left(\frac{19}{2}, 5\right)$.
8. Find the radius $r$ of the circumcircle: $r = \frac{1}{2}PQ = \frac{25}{2}$.
9. The equation of the circumcircle is $\left(x - \frac{19}{2}\right)^2 + (y - 5)^2 = \left(\frac{25}{2}\right)^2$. Since the focus $S(h, k)$ lies on this circle, substitute $(h, k)$ into this equation.
10. Two cases arise from the angle bisectors:
    • Case 1: Assume the axis is $3x - y + 14 = 0$. Then $k = 3h + 14$. Substitute this into the circumcircle equation to find $h$. This yields $h = -1/2$ or $h = -3$. $h = -3$ leads to $S(-3, 5)$, which is $R$, implying a degenerate parabola, so we discard it. Thus, $h = -1/2$, which gives $k = 25/2$. So, Focus is $\left(-\frac{1}{2}, \frac{25}{2}\right)$ and Directrix is $x + 3y - 12 = 0$.
    • Case 2: Assume the axis is $x + 3y - 12 = 0$. Then $h = 12 - 3k$. Substitute this into the circumcircle equation to find $k$. This yields $k = 5$ or $k = -5/2$. $k = 5$ leads to $S(-3, 5)$, which is $R$, so we discard it. Thus, $k = -5/2$, which gives $h = 39/2$. So, Focus is $\left(\frac{39}{2}, -\frac{5}{2}\right)$ and Directrix is $3x - y + 14 = 0$.
11. Both sets of focus/directrix define valid parabolas satisfying the given conditions.

Answer: The question asks for properties of "A KCLS parabola". Based on the analysis, there are two such parabolas. The possible foci are $\left(-\frac{1}{2}, \frac{25}{2}\right)$ and $\left(\frac{39}{2}, -\frac{5}{2}\right)$. The possible directrices are $x + 3y - 12 = 0$ and $3x - y + 14 = 0$. The intersection of the tangents is $R(-3, 5)$. The angle between the tangents is $90^\circ$.

Since no options are provided, these are the properties that can be deduced.

Subject: Mathematics Chapter: Conic Sections Topic: Parabola (Tangents and Normals, Properties of Parabola) Difficulty Level: Medium Question Type: Descriptive (or potentially multiple choice with one of the derived properties as an option)