Q.Find the equation for the ellipse that satisfies the given conditions: Centre at $(0, 0)$, major axis on the $y$-axis and passes through the points $(3, 2)$ and $(1, 6)$.

The key idea: The endpoints of the major axis tell us the ellipse is centered at the origin with its major axis along the x-axis, and the distance from the center to each endpoint is the semi-major axis length $a$. A focus lies on a given line, so we can find its coordinates using the fact that the foci are on the major axis (the x-axis). Then we use the relationship $c^2 = a^2 - b^2$ to find $b$, the semi-minor axis length.

1. Identify the center and $a$.

   The endpoints of the major axis are $(5,0)$ and $(-5,0)$. The midpoint is the center: $(0,0)$. The distance from the center to either endpoint is $a = 5$. So the ellipse is of the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a=5$, i.e., $\frac{x^2}{25} + \frac{y^2}{b^2} = 1$.

2. Locate the focus.

   The foci lie on the major axis (the x-axis), so each focus has coordinates $(\pm c, 0)$ where $c > 0$ and $c^2 = a^2 - b^2$. One focus lies on the line $3x - 5y - 9 = 0$. Since the focus is on the x-axis, $y=0$. Substituting $y=0$ into the line equation gives $3x - 9 = 0$, so $x = 3$. Thus one focus is at $(3,0)$. (The other focus would be at $(-3,0)$, but we only need $c=3$.)

3. Find $b$ using $c^2 = a^2 - b^2$. …

Eliminate the parameter. From $x=3(\cos t+\sin t)$ and $y=4(\cos t-\sin t)$,

Squaring and adding, the cross terms cancel:

Write in standard form.

$\frac{x^2}{9}+\frac{y^2}{16}=2\ \Longrightarrow\ \frac{x^2}{18}+\frac{y^2}{32}=1.$ …

We are given an ellipse and a hyperbola whose foci coincide. That means the two focal points (one on each side of the origin) are the same for both curves. Since both are centered at the origin and symmetric about both axes, their foci lie on the x‑axis. So we just need to find the x‑coordinate of the foci for each and set them equal.

1. Find the foci of the hyperbola The hyperbola is

Multiply through by 25 to get standard form:

So $a_h^2 = \frac{144}{25}$ and $b_h^2 = \frac{81}{25}$. For a hyperbola of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the distance from the center to each focus is $c_h = \sqrt{a_h^2 + b_h^2}$.

Hence the foci of the hyperbola are at $(\pm 3, 0)$.

2. Find the foci of the ellipse The ellipse is

Here $a_e^2 = 16$ (since the larger denominator is under $x^2$, the major axis is horizontal). For an ellipse, the distance from the center to each focus is $c_e = \sqrt{a_e^2 - b_e^2}$, provided $a_e^2 > b_e^2$.

So the foci of the ellipse are at $(\pm \sqrt{16 - b^2}, 0)$.

3. Set the foci equal …

The key idea is to translate the geometric condition into algebra using standard ellipse parameters. For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a > b > 0$, the distance between the foci is $2ae$, and the length of the latus rectum is $\frac{2b^2}{a}$. Setting them equal gives a relation between $a$ and $e$, which simplifies to a quadratic in $e$.

1. Recall the standard ellipse parameters

   For an ellipse centered at the origin with semi-major axis $a$ and semi-minor axis $b$, the eccentricity $e$ satisfies $b^2 = a^2(1 - e^2)$. The foci are at $(\pm ae, 0)$, so the distance between them is $2ae$. The latus rectum (a chord through a focus perpendicular to the major axis) has length $\frac{2b^2}{a}$.

2. Set the condition

   The problem states:

So:

Cancel the factor 2 (assuming $a > 0$):

3. Substitute $b^2$ in terms of $a$ and $e$ Using $b^2 = a^2(1 - e^2)$, we get:

Cancel $a$ (again $a > 0$):

4. Solve the quadratic Rearranging:

The quadratic formula gives:

$e = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2}$ …

We are told the ellipse is in standard form, meaning its axes are aligned with the coordinate axes and its center is at the origin. The general equation is:

where $a$ is the semi-major axis and $b$ the semi-minor axis. The latus rectum length depends on which axis is longer. For an ellipse with $a > b$, the latus rectum is $\frac{2b^2}{a}$; if $b > a$, it is $\frac{2a^2}{b}$. We’ll determine which case holds after finding $a^2$ and $b^2$.

1. Substitute the first point $\left(\frac{5}{2}, \frac{\sqrt{6}}{4}\right)$:

Multiply through by $8a^2b^2$ to clear denominators:

2. Substitute the second point $\left(-2, \frac{\sqrt{15}}{5}\right)$:

Multiply through by $5a^2b^2$:

3. Solve the system

We have:

Subtract (2) from (1):

Assuming $b \neq 0$, divide both sides by $3b^2$:

So $a^2 = 10$.

4. Find $b^2$ …

We start with the definition: an ellipse is the set of points whose distance from a fixed point (focus) is a constant fraction $e$ (eccentricity) of its perpendicular distance to a fixed line (directrix). Here focus $F=(1,0)$, directrix $x=4$, and $e$ is a root of $2x^2-3x+1=0$.

1. Find the eccentricity.

Solve $2x^2-3x+1=0$. Factoring: $(2x-1)(x-1)=0$, so roots are $x=1$ and $x=1/2$. For an ellipse, $0<e<1$, so we take $e=1/2$. (The root $e=1$ would give a parabola, not an ellipse.)

2. Set up the focus–directrix condition.

Let $P=(x,y)$ be any point on the ellipse. Distance to focus: $\sqrt{(x-1)^2+y^2}$.

Distance to directrix $x=4$: $|x-4|$ (perpendicular distance).

The definition says:

With $e=1/2$:

3. Square both sides (both sides non-negative).

Multiply through by 4:

4. Expand and simplify.

Left: $4(x^2-2x+1)+4y^2=4x^2-8x+4+4y^2$.

Right: $x^2-8x+16$.

Equate:

Cancel $-8x$ on both sides:

Bring terms:

Let the semi-axes be $a$ (major) and $b$ (minor). The condition on lengths gives

Eccentricity: …

Divide by $24$: $\dfrac{x^2}{8}+\dfrac{y^2}{6}=1$, an ellipse with $a^2=8>b^2=6$, so $a=2\sqrt2$. By the defining property of an ellipse, the sum of distances from any point to t …