If `x = h+ asectheta` and `y=k + b"cosec"theta` then

Options

A. 1)`a^(2)/(x+h)^(2) - b^(2)/(y+k)^(2)=01`
B. 2)`a^(2)/(x-h)^(2) + b^(2)/(y-k)^(2) =1`
C. 3)`(x-h)^(2)/a^(2) + (y-k)^(2)/b^(2) =1`
D. 4)`(x-h)^(2)/a^(2) -(y-k)^(2)/b^(2) =1`

Answer

To solve the problem step by step, we start with the given equations:

1. **Given Equations:**
\[ x = h + a \sec \theta \]
\[ y = k + b \csc \theta \]

2. **Rearranging the First Equation:**
From the first equation, we can isolate \(\sec \theta\):
\[ x - h = a \sec \theta \]
Dividing both sides by \(a\):
\[ \sec \theta = \frac{x - h}{a} \]

3. **Finding \(\cos \theta\):**
Since \(\sec \theta = \frac{1}{\cos \theta}\), we can express \(\cos \theta\) as:
\[ \cos \theta = \frac{a}{x - h} \]
This is our **Equation 1**.

4. **Rearranging the Second Equation:**
From the second equation, we isolate \(\csc \theta\):
\[ y - k = b \csc \theta \]
Dividing both sides by \(b\):
\[ \csc \theta = \frac{y - k}{b} \]

5. **Finding \(\sin \theta\):**
Since \(\csc \theta = \frac{1}{\sin \theta}\), we can express \(\sin \theta\) as:
\[ \sin \theta = \frac{b}{y - k} \]
This is our **Equation 2**.

6. **Using Pythagorean Identity:**
We know from trigonometric identities that:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
Substituting our expressions for \(\sin \theta\) and \(\cos \theta\):
\[ \left(\frac{b}{y - k}\right)^2 + \left(\frac{a}{x - h}\right)^2 = 1 \]

7. **Rearranging the Equation:**
This can be rearranged to:
\[ \frac{a^2}{(x - h)^2} + \frac{b^2}{(y - k)^2} = 1 \]

Thus, we have derived the relationship between \(x\), \(y\), \(h\), \(k\), \(a\), and \(b\).

### Final Result:
\[ \frac{a^2}{(x - h)^2} + \frac{b^2}{(y - k)^2} = 1 \]