The points at which the tangents to the curve `ax^(2) + 2hxy + by^(2) =1` is parallel to `y`-axis is

Options

A. (0,0)
B. where `hx + by = 0` meets it
C. where `ax + hy` meets it
D. none of these

Answer

To find the points at which the tangents to the curve \( ax^2 + 2hxy + by^2 = 1 \) are parallel to the y-axis, we can follow these steps:

### Step 1: Differentiate the equation implicitly
We start with the given equation of the curve:
\[ ax^2 + 2hxy + by^2 = 1 \]
To find the slope of the tangent line, we differentiate both sides with respect to \( x \).

Using implicit differentiation:
\[ \frac{d}{dx}(ax^2) + \frac{d}{dx}(2hxy) + \frac{d}{dx}(by^2) = \frac{d}{dx}(1) \]
This gives:
\[ 2ax + 2hy + 2hxy' + 2byy' = 0 \]

### Step 2: Solve for \( y' \)
Now, we can rearrange this equation to solve for \( y' \):
\[ 2hxy' + 2byy' = -2ax - 2hy \]
Factoring out \( y' \):
\[ y'(2hx + 2by) = -2ax - 2hy \]
Thus,
\[ y' = \frac{-2ax - 2hy}{2hx + 2by} \]
Simplifying, we get:
\[ y' = \frac{-(ax + hy)}{hx + by} \]

### Step 3: Set the denominator to zero for vertical tangents
For the tangent to be parallel to the y-axis, the slope \( y' \) must be undefined, which occurs when the denominator is zero:
\[ hx + by = 0 \]
This can be rearranged to find the relationship between \( x \) and \( y \):
\[ by = -hx \quad \Rightarrow \quad y = -\frac{h}{b}x \]

### Step 4: Substitute back into the original equation
Now, we substitute \( y = -\frac{h}{b}x \) back into the original curve equation to find the points:
\[ ax^2 + 2hx\left(-\frac{h}{b}x\right) + b\left(-\frac{h}{b}x\right)^2 = 1 \]
This simplifies to:
\[ ax^2 - \frac{2h^2}{b}x^2 + \frac{h^2}{b}x^2 = 1 \]
Combining like terms gives:
\[ \left(a - \frac{2h^2}{b} + \frac{h^2}{b}\right)x^2 = 1 \]
\[ \left(a - \frac{h^2}{b}\right)x^2 = 1 \]
Thus,
\[ x^2 = \frac{1}{a - \frac{h^2}{b}} \quad \Rightarrow \quad x = \pm \sqrt{\frac{1}{a - \frac{h^2}{b}}} \]

### Step 5: Find corresponding \( y \) values
Using \( y = -\frac{h}{b}x \), we can find the corresponding \( y \) values:
\[ y = -\frac{h}{b}\left(\pm \sqrt{\frac{1}{a - \frac{h^2}{b}}}\right) \]

### Final Points
The points at which the tangents to the curve are parallel to the y-axis are:
\[ \left(\sqrt{\frac{1}{a - \frac{h^2}{b}}}, -\frac{h}{b}\sqrt{\frac{1}{a - \frac{h^2}{b}}}\right) \quad \text{and} \quad \left(-\sqrt{\frac{1}{a - \frac{h^2}{b}}}, \frac{h}{b}\sqrt{\frac{1}{a - \frac{h^2}{b}}}\right) \]