a. Find the equation of the straight line that has y intercept '2' and is parallel to 4x 8y + 3 = 0 . b. Find the equation of the circle which passes through origin and makes equal intercepts on…

a. Equation of the straight line with $y$-intercept 2 and parallel to $4x-8y+3=0$

Step 1: Identify the slope of the given line. The given line is $4x-8y+3=0$. Rewrite it in slope-intercept form $y=mx+c$: $$4x-8y+3=0⟹-8y=-4x-3⟹y=\frac{1}{2}x+\frac{3}{8}.$$ Thus, the slope $m=\frac{1}{2}$.

Step 2: Use the slope and $y$-intercept to write the equation. Since the desired line is parallel, it has the same slope $m=\frac{1}{2}$. Given the $y$-intercept $c=2$, the equation is: $$y=\frac{1}{2}x+2.$$

Step 3: Convert to standard form. Multiply both sides by 2 to eliminate the fraction: $$2y=x+4⟹x-2y+4=0.$$

Final Answer: The equation of the line is $x-2y+4=0$.

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b. Equation of the circle passing through the origin and making equal intercepts on both axes

Step 1: Assume the circle’s equation in intercept form. A circle with equal intercepts $a$ on both axes has the equation: $$x^2+y^2-ax-ay=0.$$

Step 2: Use the condition that the circle passes through the origin $(0,0)$. Substitute $x=0$ and $y=0$: $$0+0-0-0=0⟹0=0.$$ This holds for any $a$, so we need another condition.

Step 3: Rewrite the equation in standard form. Complete the square for $x$ and $y$: $$x^2-ax+y^2-ay=0⟹{\left(x-\frac{a}{2}\right)}^2+{\left(y-\frac{a}{2}\right)}^2=\frac{a^2}{2}.$$ The center is at $\left(\frac{a}{2},\frac{a}{2}\right)$ and the radius is $\frac{a}{\sqrt{2}}$.

Step 4: Use the condition that the circle passes through the origin. The distance from the center to the origin equals the radius: $$\sqrt{{\left(\frac{a}{2}\right)}^2+{\left(\frac{a}{2}\right)}^2}=\frac{a}{\sqrt{2}}⟹\sqrt{\frac{a^2}{4}+\frac{a^2}{4}}=\frac{a}{\sqrt{2}}⟹\sqrt{\frac{a^2}{2}}=\frac{a}{\sqrt{2}}.$$ This simplifies to $\frac{|a|}{\sqrt{2}}=\frac{a}{\sqrt{2}}$, which holds for all $a\ne 0$.

Step 5: Simplify the equation. Since the circle passes through the origin and has equal intercepts, the simplest form is when $a=2r$ (where $r$ is the radius). However, the general form is: $$x^2+y^2-ax-ay=0.$$ For the circle to pass through the origin and have equal intercepts, the equation reduces to: $$x^2+y^2-2rx-2ry=0.$$ But the most straightforward solution is when the circle’s equation is symmetric and passes through $(0,0)$. Testing $a=2$ gives: $$x^2+y^2-2x-2y=0.$$

Verification:

• Intercepts: Set $y=0$, $x^2-2x=0⟹x=0,2$. Set $x=0$, $y^2-2y=0⟹y=0,2$. Thus, equal intercepts of 2 on both axes.

Final Answer: The equation of the circle is $x^2+y^2-2x-2y=0$.

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c. Evaluate: ${\lim }_{x\to 0}\frac{1-\cos mx}{1-\cos nx}$

Step 1: Apply the trigonometric identity $1-\cos \theta =2{\sin }^2\frac{\theta }{2}$: $$\underset{x\to 0}{\lim }\frac{2{\sin }^2\frac{mx}{2}}{2{\sin }^2\frac{nx}{2}}=\underset{x\to 0}{\lim }\frac{{\sin }^2\frac{mx}{2}}{{\sin }^2\frac{nx}{2}}.$$

Step 2: Use the small-angle approximation $\sin \theta \approx \theta$ for $\theta \to 0$: $$\sin \frac{mx}{2}\approx \frac{mx}{2},\sin \frac{nx}{2}\approx \frac{nx}{2}.$$ Thus: $$\frac{{\sin }^2\frac{mx}{2}}{{\sin }^2\frac{nx}{2}}\approx \frac{{\left(\frac{mx}{2}\right)}^2}{{\left(\frac{nx}{2}\right)}^2}=\frac{m^2{x}^2/4}{n^2{x}^2/4}=\frac{m^2}{n^2}.$$

Step 3: Take the limit as $x\to 0$: $$\underset{x\to 0}{\lim }\frac{m^2}{n^2}=\frac{m^2}{n^2}.$$

Final Answer: The limit evaluates to $\frac{m^2}{n^2}$.