Solve question 5,6 in English language

Here are the solutions for questions 5 and 6.

Question 5:

Part 1: Show that the curve is a parabola. The given polar equation is $r = \frac{2}{4 \cos\theta + 1}$. To identify the conic section, we rewrite the equation in the standard form $r = \frac{ed}{1 + e \cos\theta}$. Divide the numerator and denominator by 4: $r = \frac{\frac{2}{4}}{\frac{4 \cos\theta}{4} + \frac{1}{4}} = \frac{\frac{1}{2}}{1 + \frac{1}{4} \cos\theta}$ Comparing this to the standard form $r = \frac{ed}{1 + e \cos\theta}$, we identify the eccentricity $e = \frac{1}{4}$. Since $e < 1$, the curve is an ellipse, not a parabola.

Correction: I made a mistake in my thought process. The given equation is $r = \frac{2}{4 \cos\theta + 1}$. The standard form for a conic section is $r = \frac{ed}{1 \pm e \cos\theta}$. To match this, we need the denominator to start with 1. $r = \frac{2}{1 + 4 \cos\theta}$ Here, $e=4$ and $ed=2$, so $d = \frac{2}{4} = \frac{1}{2}$. Since $e=4 > 1$, the curve is a hyperbola, not a parabola.

There seems to be a discrepancy in the question. If the question states "Show that the curve... is a parabola", but the equation given ($r = \frac{2}{4 \cos\theta + 1}$) represents a hyperbola ($e=4$), then the premise of the question is incorrect.

However, if the question intended the equation to be $r = \frac{2}{1 + \cos\theta}$ (which would be a parabola with $e=1$), or if the question meant $r = \frac{2}{4 \cos\theta + 4}$ (which would be $r = \frac{1/2}{1 + \cos\theta}$, also a parabola), then the problem would make sense.

Given the exact equation $r = \frac{2}{4 \cos\theta + 1}$: The eccentricity is $e=4$. Since $e > 1$, the curve is a hyperbola. Therefore, the statement "the curve... is a parabola" is incorrect for the given equation.

I will proceed assuming there might be a typo in the question and that it intended to be a parabola, or I will state that it is a hyperbola. Given the instruction "Show that the curve... is a parabola", I must point out the contradiction.

Let's re-evaluate the image. The equation is $r = \frac{2}{4 \cos\theta + 1}$. This is clear. The statement "Is a parabola, P, in the (x, y) plane." is also clear. This is a direct contradiction. I must state this.

Part 1: Analysis of the curve. The given polar equation is $r = \frac{2}{4 \cos\theta + 1}$. To determine the type of conic section, we compare it to the standard form $r = \frac{ed}{1 + e \cos\theta}$. We can rewrite the given equation as: $r = \frac{2}{1 + 4 \cos\theta}$ By comparing, we find that the eccentricity $e = 4$. Since $e > 1$, the curve is a hyperbola, not a parabola. Therefore, the premise of the question, "Show that the curve... is a parabola", is incorrect for the given equation.

Part 2: Show that the point $(0, \frac{1}{2})$ lies on $P$. Let's convert the polar equation to Cartesian coordinates. From $r = \frac{2}{1 + 4 \cos\theta}$: $r(1 + 4 \cos\theta) = 2$ $r + 4r \cos\theta = 2$ Substitute $r = \sqrt{x^2 + y^2}$ and $x = r \cos\theta$: $\sqrt{x^2 + y^2} + 4x = 2$ $\sqrt{x^2 + y^2} = 2 - 4x$ Square both sides: $x^2 + y^2 = (2 - 4x)^2$ $x^2 + y^2 = 4 - 16x + 16x^2$ $y^2 = 15x^2 - 16x + 4$. This is the Cartesian equation of the hyperbola.

Now, we check if the point $(0, \frac{1}{2})$ lies on this curve. Substitute $x=0$ and $y=\frac{1}{2}$: $\left(\frac{1}{2}\right)^2 = 15(0)^2 - 16(0) + 4$ $\frac{1}{4} = 0 - 0 + 4$ $\frac{1}{4} = 4$ This statement is false. Therefore, the point $(0, \frac{1}{2})$ does not lie on the curve $P$ defined by $r = \frac{2}{4 \cos\theta + 1}$.

Part 3: Find the equation of the tangent to $P$ at the point $(0, \frac{1}{2})$. Since the point $(0, \frac{1}{2})$ does not lie on the curve, it is not possible to find a tangent to the curve at this point.

Given the contradictions in the question (the curve is a hyperbola, not a parabola, and the point does not lie on the curve), it is likely there is a typo in the problem statement. If the question intended a parabola, the equation would need to be different (e.g., $r = \frac{2}{1 + \cos\theta}$ or $r = \frac{2}{2 + 2 \cos\theta}$). If we assume the question meant $r = \frac{2}{2 + 2 \cos\theta} = \frac{1}{1 + \cos\theta}$ (which is a parabola), then the point $(0, \frac{1}{2})$ would lie on it. However, I must solve the problem as written.

Conclusion for Question 5: Based on the given equation $r = \frac{2}{4 \cos\theta + 1}$:

• The curve is a hyperbola (eccentricity $e=4$), not a parabola.
• The point $(0, \frac{1}{2})$ does not lie on this curve.
• Therefore, finding the equation of the tangent at this point is not possible.

Question 6 (a):

We need to solve the system of congruences using the Chinese Remainder Theorem: $x \equiv 3 \pmod{7}$ $x \equiv 5 \pmod{8}$

Step 1: Identify the parameters. We have $a_1 = 3$, $m_1 = 7$. And $a_2 = 5$, $m_2 = 8$.

Step 2: Check if the moduli are coprime. The moduli are $m_1 = 7$ and $m_2 = 8$. $\text{gcd}(7, 8) = 1$, so they are coprime.

Step 3: Calculate $M$. $M = m_1 \times m_2 = 7 \times 8 = 56$.

Step 4: Calculate $M_i$. $M_1 = \frac{M}{m_1} = \frac{56}{7} = 8$. $M_2 = \frac{M}{m_2} = \frac{56}{8} = 7$.

Step 5: Find the modular inverses $y_i$. For $M_1 y_1 \equiv 1 \pmod{m_1}$: $8 y_1 \equiv 1 \pmod{7}$ Since $8 \equiv 1 \pmod{7}$, this simplifies to $1 y_1 \equiv 1 \pmod{7}$. So, $y_1 = 1$.

For $M_2 y_2 \equiv 1 \pmod{m_2}$: $7 y_2 \equiv 1 \pmod{8}$ We can write $7 \equiv -1 \pmod{8}$, so $-1 y_2 \equiv 1 \pmod{8}$. Multiplying by $-1$, we get $y_2 \equiv -1 \pmod{8}$. The smallest positive integer for $y_2$ is $y_2 = 7$.

Step 6: Construct the solution for $x$. The solution is given by $x \equiv (a_1 M_1 y_1 + a_2 M_2 y_2) \pmod{M}$. $x \equiv (3 \times 8 \times 1 + 5 \times 7 \times 7) \pmod{56}$ $x \equiv (24 + 245) \pmod{56}$ $x \equiv 269 \pmod{56}$

Step 7: Simplify the solution. To find the smallest non-negative integer solution, we divide 269 by 56: $269 = 4 \times 56 + 45$ So, $269 \equiv 45 \pmod{56}$.

The solution is $\boxed{x \equiv 45 \pmod{56}}$.