Q.Find the coordinates of the foci and the vertices, the eccentricity, the length of the latus rectum of the hyperbolas:

Concept & Intuition

The problem ties together two conic sections that share the same foci. For an ellipse, $c^2 = a^2 - b^2$; for a hyperbola, $c^2 = a^2 + b^2$. The key is to first find the ellipse’s foci, then use the hyperbola’s given transverse axis length to determine its $a$, and finally solve for $b^2$ using the shared $c$.

Step-by-step solution

1. Rewrite the ellipse in standard form The ellipse equation is $9x^2 + 16y^2 = 144$. Divide through by 144:

So $a_{\text{ell}}^2 = 16$, $b_{\text{ell}}^2 = 9$. Since $16 > 9$, the major axis is horizontal.

2. Find the foci of the ellipse

   For an ellipse, $c^2 = a^2 - b^2 = 16 - 9 = 7$, so $c = \sqrt{7}$.

   The foci are at $(\pm \sqrt{7}, 0)$.

3. The hyperbola shares these foci

   Therefore, for the hyperbola, $c = \sqrt{7}$ as well, and its foci are also $(\pm \sqrt{7}, 0)$. This tells us the hyperbola’s transverse axis is horizontal, so its equation will be of the form

4. Use the given transverse axis length The length of the transverse axis is $2a = 2\cos\alpha$, so

Hence $a^2 = \cos^2\alpha$.

5. Relate $a$, $b$, and $c$ for a hyperbola For a hyperbola, $c^2 = a^2 + b^2$. Substituting $c^2 = 7$ and $a^2 = \cos^2\alpha$:

6. Write the hyperbola’s equation Plug $a^2$ and $b^2$ into the standard form:

7. Match with the options This matches option (B) exactly.

Concept & Intuition

For a hyperbola, the defining property is that the absolute difference of distances from any point to the two foci is constant and equal to $2a$, where $a$ is the semi-transverse axis. The eccentricity $e = c/a$, where $c$ is the distance from the center to each focus. The relationship $c^2 = a^2 + b^2$ links the transverse and conjugate axes. Once we find $a$ and $b$, the standard equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ (for a horizontal transverse axis) is determined.

Step-by-step solution

1. Identify $a$ from the given difference

   The problem states: “The difference between the distance of any point on the hyperbola from the two foci is 16.”

   By definition, this constant difference is $2a$.

   So $2a = 16 \implies a = 8$.

2. Use eccentricity to find $c$

   Eccentricity $e = 2$ is given. For a hyperbola, $e = \frac{c}{a}$.

   Hence $c = a e = 8 \times 2 = 16$.

3. Find $b^2$ using the hyperbola relation

   For a hyperbola, $c^2 = a^2 + b^2$.

   Substitute: $16^2 = 8^2 + b^2 \implies 256 = 64 + b^2 \implies b^2 = 192$.

4. Write the equation

   With $a^2 = 64$ and $b^2 = 192$, the standard form (assuming the transverse axis is along the $x$-axis, which is the usual default) is

5. Match with the options This corresponds exactly to option (C).

Concept & Intuition

The latus rectum of a conic is a chord through a focus perpendicular to the major (or transverse) axis. For a hyperbola, its length is a fixed geometric property derived from the standard equation. The key is to first identify the type of conic and then put it into standard form so we can read off the parameters $a$ and $b$. Here, the equation has a positive $y^2$ term and a negative $x^2$ term, so it’s a hyperbola opening upward and downward (vertical transverse axis). The formula for the length of the latus rectum for such a hyperbola is $\frac{2b^2}{a}$, where $a$ is the distance from the center to a vertex along the transverse axis, and $b$ relates to the conjugate axis.

Step-by-step solution

1. Rewrite the equation in standard form Start with $49y^2 - 16x^2 = 784$. Divide both sides by 784 to get 1 on the right:

Simplify each fraction:

This is the standard form of a hyperbola with a vertical transverse axis:

where $a^2 = 16$ and $b^2 = 49$.

2. Identify $a$ and $b$

   From $a^2 = 16$, we get $a = 4$ (positive length).

   From $b^2 = 49$, we get $b = 7$.

3. Recall the latus rectum formula for a hyperbola

   For a hyperbola of the form $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$, the length of the latus rectum is:

This is derived from the fact that the latus rectum passes through a focus at $(0, \pm c)$ where $c^2 = a^2 + b^2$, and its endpoints satisfy the hyperbola equation with $y = c$.

4. Plug in the values

Concept & Intuition

For a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the foci are at $(\pm ae, 0)$ and the directrices are the vertical lines $x = \pm a/e$.

The distance between the foci is $2ae$, and the distance between the two directrices is $2a/e$.

The problem gives the ratio of these two distances as $3:2$, which lets us solve for the eccentricity $e$. Once we have $e$, we use the relation $b^2 = a^2(e^2 - 1)$ to find the ratio $a:b$.

Step-by-step solution

1. Write the given ratio Distance between foci: $2ae$ Distance between directrices: $2a/e$ Their ratio is $3:2$, so:

The $2a$ cancels, leaving:

Hence:

2. Relate $b$ to $a$ using eccentricity For a hyperbola, $e^2 = 1 + \frac{b^2}{a^2}$. Substitute $e^2 = 3/2$:

So:

3. Find $a:b$ From $\frac{b^2}{a^2} = \frac{1}{2}$, take square roots (positive lengths):

Therefore $a:b = \sqrt{2}:1$.

The distance between the foci is $2c=16$, so $c=8$.

Eccentricity $e=\dfrac{c}{a}=\sqrt2$ gives $a=\dfrac{c}{\sqrt2}=\dfrac{8}{\sqrt2}=4\sqrt2$, hence $a^2=32$.

For a hyperbola $b^2=c^2-a^2=64-32=32$.

Equation: $\dfrac{x^2}{32}-\dfrac{y^2}{32}=1$, i.e. $x^2-y^2=32$ (a rectangular hyperbola, consistent with $e=\sqrt2$).