Q.Find the equation of the hyperbola with:

Here are the equations of the hyperbolas based on the given conditions:

(a) Vertices $(\pm 5, 0)$, foci $(\pm 7, 0)$

• Concept: Standard form of a horizontal hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
• Reasoning:
  1. From vertices $(\pm 5, 0)$, we have $a = 5$, so $a^2 = 25$.
  2. From foci $(\pm 7, 0)$, we have $c = 7$, so $c^2 = 49$.
  3. Using the relation $c^2 = a^2 + b^2$, we find $49 = 25 + b^2$, which gives $b^2 = 24$. …

The equation of a hyperbola depends on the orientation of its transverse axis and the values of its semi-transverse axis ($a$) and semi-conjugate axis ($b$). The distance from the center to each focus is $c$. These parameters are related by $c^2 = a^2 + b^2$.

There are two standard forms for a hyperbola centered at the origin:

1. Transverse axis along the x-axis: Foci are $(\pm c, 0)$ and vertices are $(\pm a, 0)$. The equation is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
2. Transverse axis along the y-axis: Foci are $(0, \pm c)$ and vertices are $(0, \pm a)$. The equation is $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$.

Let's solve each part using these concepts.

(a) Vertices $(\pm 5, 0)$, foci $(\pm 7, 0)$

1. Identify the orientation and parameters:

   The vertices are $(\pm 5, 0)$ and the foci are $(\pm 7, 0)$. Since both lie on the x-axis, the transverse axis of the hyperbola is along the x-axis.

   From the definition of vertices, $a = 5$, so $a^2 = 25$.

   From the definition of foci, $c = 7$, so $c^2 = 49$.

2. Find $b^2$ using the fundamental relation:

   The relationship between $a$, $b$, and $c$ for a hyperbola is $c^2 = a^2 + b^2$.

   Substitute the values of $a^2$ and $c^2$:

   $49 = 25 + b^2$

   $b^2 = 49 - 25$

   $b^2 = 24$

3. Write the equation of the hyperbola:

   Since the transverse axis is along the x-axis, the standard form is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

   Substitute $a^2 = 25$ and $b^2 = 24$:

   $\frac{x^2}{25} - \frac{y^2}{24} = 1$

(b) Vertices $(0, \pm 7)$, $e = \dfrac{4}{3}$

1. Identify the orientation and parameters:

   The vertices are $(0, \pm 7)$. Since they lie on the y-axis, the transverse axis of the hyperbola is along the y-axis.

   From the definition of vertices, $a = 7$, so $a^2 = 49$.

   The eccentricity is given as $e = \frac{4}{3}$.

2. Find $c$ using eccentricity:

   The eccentricity of a hyperbola is defined as $e = \frac{c}{a}$.

   We have $e = \frac{4}{3}$ and $a = 7$.

   $\frac{4}{3} = \frac{c}{7}$

   $c = \frac{4 \times 7}{3} = \frac{28}{3}$

   So, $c^2 = \left(\frac{28}{3}\right)^2 = \frac{784}{9}$.

3. Find $b^2$ using the fundamental relation:

   Use the relation $c^2 = a^2 + b^2$.

   Substitute $a^2 = 49$ and $c^2 = \frac{784}{9}$:

   $\frac{784}{9} = 49 + b^2$

   $b^2 = \frac{784}{9} - 49$

   To subtract, find a common denominator: $49 = \frac{49 \times 9}{9} = \frac{441}{9}$.

   $b^2 = \frac{784}{9} - \frac{441}{9} = \frac{784 - 441}{9} = \frac{343}{9}$

4. Write the equation of the hyperbola:

   Since the transverse axis is along the y-axis, the standard form is $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$.

   Substitute $a^2 = 49$ and $b^2 = \frac{343}{9}$:

   $\frac{y^2}{49} - \frac{x^2}{343/9} = 1$

   This can be rewritten as:

   $\frac{y^2}{49} - \frac{9x^2}{343} = 1$

(c) Foci $(0, \pm\sqrt{10})$, passing through $(2, 3)$

1. Identify the orientation and parameters:

   The foci are $(0, \pm\sqrt{10})$. Since they lie on the y-axis, the transverse axis of the hyperbola is along the y-axis.

   From the definition of foci, $c = \sqrt{10}$, so $c^2 = 10$.

   The hyperbola passes through the point $(2, 3)$.

2. Relate $a^2$ and $b^2$ using $c^2$:

   The fundamental relation is $c^2 = a^2 + b^2$.

   Substitute $c^2 = 10$:

   $10 = a^2 + b^2$

   This means $b^2 = 10 - a^2$.

   Watch out

   Remember that $a^2$ and $b^2$ must both be positive. This implies $a^2 < 10$.

3. Form the equation using the passing point: …