Tangent and Normal Questions in English

(C) The equation of the curve is $xy = 4$, which can be written as $y = \frac{4}{x}$.
Taking the derivative with respect to $x$, we get $\frac{dy}{dx} = -\frac{4}{x^2}$.
The equation of the line is $ax + by + c = 0$, which can be rewritten as $y = -\frac{a}{b}x - \frac{c}{b}$.
The slope of this line is $m = -\frac{a}{b}$.
Since the line is a tangent to the curve, the slope of the tangent at any point $(x, y)$ on the curve must equal the slope of the line:
$-\frac{4}{x^2} = -\frac{a}{b} \implies \frac{a}{b} = \frac{4}{x^2}$.
Since $x^2 > 0$ for all $x \neq 0$, it follows that $\frac{a}{b} > 0$.
This implies that $a$ and $b$ must have the same sign.
Looking at the options, the condition $a < 0$ and $b < 0$ satisfies $\frac{a}{b} > 0$.

(C) To find the angle between the curves $y^2=x$ and $x^2=y$ at the origin $(0,0)$:
$1$. For the curve $y^2=x$, differentiating with respect to $x$ gives $2y \frac{dy}{dx} = 1$, so $\frac{dy}{dx} = \frac{1}{2y}$. At the origin $(0,0)$, the slope is undefined, which means the tangent is the $y$-axis $(x=0)$.
$2$. For the curve $x^2=y$, differentiating with respect to $x$ gives $2x = \frac{dy}{dx}$. At the origin $(0,0)$, the slope is $\frac{dy}{dx} = 0$, which means the tangent is the $x$-axis $(y=0)$.
$3$. The angle between the $x$-axis and the $y$-axis is $\frac{\pi}{2}$ radians. Therefore, the angle between the two curves at the origin is $\frac{\pi}{2}$.

(B) Given the curve $9y^2 = x^3$. Differentiating with respect to $x$, we get $18y \frac{dy}{dx} = 3x^2$, so $\frac{dy}{dx} = \frac{x^2}{6y}$.
The slope of the normal at point $(x_1, y_1)$ is $m = -\frac{1}{dy/dx} = -\frac{6y_1}{x_1^2}$.
Since the normal makes equal intercepts with the axes, its slope must be $\pm 1$. Given the curve, the normal must have a slope of $-1$ to make equal intercepts (i.e., $y - y_1 = -1(x - x_1) \implies x + y = x_1 + y_1$).
Setting $-\frac{6y_1}{x_1^2} = -1$, we get $x_1^2 = 6y_1$.
Substitute $y_1 = \frac{x_1^2}{6}$ into the curve equation $9y_1^2 = x_1^3$:
$9(\frac{x_1^2}{6})^2 = x_1^3 \implies 9(\frac{x_1^4}{36}) = x_1^3 \implies \frac{x_1^4}{4} = x_1^3$.
Since $x_1 \neq 0$, $x_1 = 4$. Then $y_1^2 = \frac{4^3}{9} = \frac{64}{9}$, so $y_1 = \pm 8/3$.
Checking the slope condition: if $x_1 = 4, y_1 = 8/3$, slope $m = -\frac{6(8/3)}{16} = -1$. Thus, the point is $(4, 8/3)$.

(D) Step $1$: Since the point $(2, 3)$ lies on the curve $y^2 = ax^3 + b$, we have $3^2 = a(2)^3 + b$, which simplifies to $9 = 8a + b$ (Equation $1$).
Step $2$: Differentiate the curve equation with respect to $x$: $2y \frac{dy}{dx} = 3ax^2$, so $\frac{dy}{dx} = \frac{3ax^2}{2y}$.
Step $3$: The slope of the tangent at $(2, 3)$ is $\frac{3a(2)^2}{2(3)} = \frac{12a}{6} = 2a$. Given the line $y = 4x - 5$ has a slope of $4$, we equate $2a = 4$, giving $a = 2$.
Step $4$: Substitute $a = 2$ into Equation $1$: $9 = 8(2) + b \implies 9 = 16 + b \implies b = -7$.
Step $5$: Calculate $7a - 2b = 7(2) - 2(-7) = 14 + 14 = 28$.

(D) Given the curve $y = \sqrt{9 - 3x^2}$. At the point where abscissa $(x)$ and ordinate $(y)$ are equal, $x = y$.
Substitute $y = x$ into the equation: $x = \sqrt{9 - 3x^2} \implies x^2 = 9 - 3x^2 \implies 4x^2 = 9 \implies x^2 = \frac{9}{4} \implies x = \frac{3}{2}$ (since $y$ must be positive).
So, the point of tangency is $(\frac{3}{2}, \frac{3}{2})$.
Differentiating $y^2 = 9 - 3x^2$ with respect to $x$: $2y \frac{dy}{dx} = -6x \implies \frac{dy}{dx} = -\frac{3x}{y}$.
At $(\frac{3}{2}, \frac{3}{2})$, the slope $m = -\frac{3(3/2)}{3/2} = -3$.
The equation of the tangent is $y - y_1 = m(x - x_1) \implies y - \frac{3}{2} = -3(x - \frac{3}{2})$.
$y - \frac{3}{2} = -3x + \frac{9}{2} \implies 3x + y - 6 = 0$.

(A) Given the curve $x = a \sin^3 t$ and $y = b \cos^3 t$.
Find the derivatives: $\frac{dx}{dt} = 3a \sin^2 t \cos t$ and $\frac{dy}{dt} = -3b \cos^2 t \sin t$.
The slope of the tangent is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-3b \cos^2 t \sin t}{3a \sin^2 t \cos t} = -\frac{b}{a} \cot t$.
At $t = \frac{\pi}{2}$, $\frac{dy}{dx} = -\frac{b}{a} \cot(\frac{\pi}{2}) = 0$.
The slope of the normal is $-\frac{1}{dy/dx}$, which is undefined (vertical line).
However, at $t = \frac{\pi}{2}$, the point is $(x, y) = (a \sin^3(\frac{\pi}{2}), b \cos^3(\frac{\pi}{2})) = (a, 0)$.
$A$ vertical line passing through $(a, 0)$ is $x = a$, or $x - a = 0$.
Comparing $x - a = 0$ with $x + By + C = 0$, we get $B = 0$ and $C = -a$.
Therefore, $B - C = 0 - (-a) = a$.

(A) $1$. The curve is $y = \frac{1}{x}$. The slope of the tangent at any point $(x_1, y_1)$ is $\frac{dy}{dx} = -\frac{1}{x_1^2} = -y_1^2$.
$2$. The slope of the normal at $(x_1, y_1)$ is $m_n = -\frac{1}{dy/dx} = \frac{1}{y_1^2} = x_1^2$.
$3$. The equation of the normal at $(x_1, y_1)$ is $y - y_1 = x_1^2(x - x_1)$, which simplifies to $x_1^2 x - y + (y_1 - x_1^3) = 0$.
$4$. Comparing this with $ax + by + 5 = 0$, we have $\frac{x_1^2}{a} = \frac{-1}{b} = \frac{y_1 - x_1^3}{5}$.
$5$. Since $x_1^2 > 0$ and $a = -b x_1^2$, for the normal to exist, $a$ and $b$ must have opposite signs. Specifically, since $x_1^2 > 0$, $a$ and $b$ must satisfy $a/b = -x_1^2 < 0$. Thus, $a$ and $b$ have opposite signs. Given the options, $a > 0$ and $b < 0$ is the correct condition.

(C) Step $1$: Since the point $(1, -1)$ lies on the curve $y^2 = ax^3 + b$, we have $(-1)^2 = a(1)^3 + b$, which simplifies to $a + b = 1$.
Step $2$: Differentiate the curve equation with respect to $x$: $2y \frac{dy}{dx} = 3ax^2$.
Step $3$: At the point $(1, -1)$, the slope of the tangent is $\frac{dy}{dx} = \frac{3a(1)^2}{2(-1)} = -\frac{3a}{2}$.
Step $4$: The line $x + y = 0$ can be written as $y = -x$, which has a slope of $-1$.
Step $5$: Equating the slopes: $-\frac{3a}{2} = -1$, which gives $a = 2/3$.
Step $6$: Substitute $a = 2/3$ into $a + b = 1$: $2/3 + b = 1$, so $b = 1/3$.

(C) Step $1$: Find the $y$-coordinate at $x = 1$. $y = 3(1)^3 - 3(1)^2 + 1 = 3 - 3 + 1 = 1$. The point of tangency is $(1, 1)$.
Step $2$: Find the slope of the tangent by differentiating $y$ with respect to $x$. $\frac{dy}{dx} = 9x^2 - 6x + 1$.
Step $3$: Evaluate the slope $m$ at $x = 1$. $m = 9(1)^2 - 6(1) + 1 = 9 - 6 + 1 = 4$.
Step $4$: Use the point-slope form $y - y_1 = m(x - x_1)$ to find the equation. $y - 1 = 4(x - 1)$.
Step $5$: Simplify the equation. $y - 1 = 4x - 4 \implies 4x - y - 3 = 0$.

(D) tangent is vertical when the slope $\frac{dy}{dx} = \infty$, which is equivalent to $\frac{dx}{dy} = 0$.
Differentiating the equation $y^2 - xy + 9 = 0$ with respect to $y$:
$\frac{d}{dy}(y^2) - \frac{d}{dy}(xy) + \frac{d}{dy}(9) = 0$
$2y - (x \cdot 1 + y \cdot \frac{dx}{dy}) = 0$
For a vertical tangent, substitute $\frac{dx}{dy} = 0$:
$2y - x = 0 \implies x = 2y$
Substitute $x = 2y$ into the original curve equation:
$y^2 - (2y)y + 9 = 0$
$y^2 - 2y^2 + 9 = 0$
$-y^2 + 9 = 0$
$y^2 = 9$
$y = \pm 3$.

(A) Given the curve $xy + 7 = 0$, we have $y = -\frac{7}{x}$.
Differentiating with respect to $x$, we get $\frac{dy}{dx} = \frac{7}{x^2}$.
The slope of the tangent at any point $(x_0, y_0)$ is $m_t = \frac{7}{x_0^2}$.
The slope of the normal $m_n$ is $-\frac{1}{m_t} = -\frac{x_0^2}{7}$.
The equation of the normal at $(x_0, y_0)$ is $y - y_0 = -\frac{x_0^2}{7}(x - x_0)$.
Multiplying by $7$, we get $7y - 7y_0 = -x_0^2 x + x_0^3$.
Rearranging terms: $x_0^2 x + 7y - (7y_0 + x_0^3) = 0$.
Comparing this with $Ax + By + C = 0$, we identify $A = x_0^2$ and $B = 7$.
Since $x_0^2 > 0$ for any $x_0 \neq 0$ and $7 > 0$, both $A$ and $B$ must have the same sign (positive).
Thus, $A > 0$ and $B > 0$.

(C) Differentiate $2y^3 = x^3 + ax^2$ with respect to $x$: $6y^2 \frac{dy}{dx} = 3x^2 + 2ax$.
At the point $(a, a)$, the slope $m = \frac{dy}{dx} = \frac{3a^2 + 2a^2}{6a^2} = \frac{5a^2}{6a^2} = \frac{5}{6}$.
The equation of the tangent at $(a, a)$ is $y - a = \frac{5}{6}(x - a)$.
Multiplying by $6$, we get $6y - 6a = 5x - 5a$, which simplifies to $5x - 6y = -a$.
To find the intercepts, rewrite the equation as $\frac{x}{-a/5} + \frac{y}{a/6} = 1$.
Thus, the intercepts are $\alpha = -a/5$ and $\beta = a/6$.
Given $\alpha^2 + \beta^2 = 61$, we have $\frac{a^2}{25} + \frac{a^2}{36} = 61$.
$\frac{36a^2 + 25a^2}{900} = 61 \implies \frac{61a^2}{900} = 61$.
$a^2 = 900 \implies a = \pm 30$.

(D) The given line is $2x - 2y = 3$, which can be written as $y = x - 1.5$. The slope of this line is $m = 1$.
Since the normal is parallel to this line, the slope of the normal $m_n = 1$.
The slope of the tangent $m_t$ is given by $-1 / m_n = -1 / 1 = -1$.
For the curve $y = x \log x$, the derivative is $\frac{dy}{dx} = \log x + x \cdot \frac{1}{x} = \log x + 1$.
Equating the slope of the tangent to the derivative: $\log x + 1 = -1$, which gives $\log x = -2$.
Thus, $x = e^{-2}$.
Substituting $x = e^{-2}$ into the curve equation: $y = e^{-2} \log(e^{-2}) = e^{-2} (-2) = -2e^{-2}$.
Therefore, the point is $(e^{-2}, -2e^{-2})$.