Q.Show that the path of a moving point such that its distances from two lines $3x - 2y = 5$ and $3x + 2y = 5$ are equal is a straight line.

The problem describes a hyperbola whose foci lie on the X-axis. This means the transverse axis (major axis) of the hyperbola is along the X-axis. The general equation for such a hyperbola, centered at $(h, k)$, is $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$. Since the foci lie on the X-axis, the $y$-coordinate of the center must be $0$, so $k=0$. The foci are located at $(h \pm ae, 0)$.

The "chord of this hyperbola which passes through its focus and perpendicular to its major axis" is precisely the definition of the latus rectum of the hyperbola. For a hyperbola with its transverse axis along the X-axis, the length of the latus rectum is given by the formula $\frac{2b^2}{a}$. Our goal is to find $a$ and $b^2$.

The eccentricity $e$ of a hyperbola is related to $a$ and $b$ by the equation $b^2 = a^2(e^2 - 1)$.

Here's how we solve the problem:

1. Determine the coordinates of the foci:

   The X-coordinates of the foci are the roots of the quadratic equation $x^2 - 4x + 1 = 0$. Let these roots be $x_1$ and $x_2$.

   From Vieta's formulas, for a quadratic equation $Ax^2 + Bx + C = 0$:

   Sum of roots: $x_1 + x_2 = -\frac{B}{A} = -\frac{-4}{1} = 4$.

   Product of roots: $x_1 x_2 = \frac{C}{A} = \frac{1}{1} = 1$.

   The foci of the hyperbola are $(x_1, 0)$ and $(x_2, 0)$.

   The center of the hyperbola $(h, 0)$ is the midpoint of the foci.

   So, $h = \frac{x_1 + x_2}{2} = \frac{4}{2} = 2$.

   The distance from the center to each focus is $ae$. This distance is half the distance between the two foci.

   $2ae = |x_2 - x_1|$.

   We can find $|x_2 - x_1|$ using the identity $(x_2 - x_1)^2 = (x_1 + x_2)^2 - 4x_1 x_2$. …

We are given:

We want the Cartesian equation satisfied by all points $(x, y)$ in $S$.

Concept and intuition:

The coordinates are linear combinations of $\cos t$ and $\sin t$. This suggests we can treat $\cos t$ and $\sin t$ as unknowns, solve for them in terms of $x$ and $y$, and then use the fundamental identity $\cos^2 t + \sin^2 t = 1$ to eliminate $t$. The result will be a quadratic equation in $x$ and $y$ — a conic section.

Let’s proceed step by step.

1. Set up the system Write the parametric equations as:

We treat $\cos t$ and $\sin t$ as unknowns.

2. Solve for $\cos t$ and $\sin t$ Subtract the second equation from the first? Better: solve the linear system. Multiply the first equation by 3 and the second by 2, then subtract:

So:

Similarly, multiply the first by 2 and the second by 3, then subtract:

So:

3. Apply the Pythagorean identity Since $\cos^2 t + \sin^2 t = 1$, substitute:

4. Simplify the equation Multiply through by 25: $(3y - 2x)^2 + (3x - 2y)^2 = 25.$ …

Concept and intuition

A circle “inscribed” in a parabola means it is tangent to the parabola — it touches it without crossing. Because the parabola is symmetric about the x‑axis, any such circle that also passes through the focus $(a,0)$ must be symmetric too, so its center lies on the x‑axis. The key idea: the circle and parabola share a tangent at the point of contact, and the distance from the circle’s center to the focus equals the radius. We use the standard parametric form of the parabola to find the contact points, then impose tangency and the focus condition.

Step‑by‑step solution

1. Set up the parabola and circle The parabola is $y^2 = 4ax$ with focus $S = (a,0)$. Let the circle have center $C = (h,0)$ on the x‑axis (by symmetry) and radius $r$. Its equation is

2. Parametrize a point on the parabola

   Any point on $y^2 = 4ax$ can be written as $P = (at^2, 2at)$ for a real parameter $t$. Because the circle is inscribed, it touches the parabola at two symmetric points (one for $t$ and one for $-t$). We’ll work with $t > 0$ for one contact point.

3. Condition for tangency

   At the point of contact $P$, the circle and parabola have the same tangent line. The slope of the tangent to the parabola at $P$ is

The radius $CP$ is perpendicular to this tangent, so its slope is $-t$. Hence the vector from $C$ to $P$ has slope $-t$:

Simplify:

4. Use the focus condition The circle passes through the focus $S = (a,0)$, so the distance $CS = r$:

Also, the radius equals the distance from $C$ to $P$:

Substitute $h = at^2 + 2a$ from step 3 into the expression for $r$:

Setup. Write the ellipse in standard form:

Then $c^2=a^2-b^2=1$, so $c=1$ and the latus-rectum ends are at $x=\pm 1,\; y=\pm\dfrac{b^2}{a}=\pm\dfrac{3}{2}$.

The end in the third quadrant is $L_1'=\left(-1,-\dfrac{3}{2}\right)$.

Normal at $L_1'$. The normal to $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ at $(x_0,y_0)$ is

Substituting $a^2=4,\,b^2=3,\,x_0=-1,\,y_0=-\dfrac{3}{2}$:

$\frac{4x}{-1}-\frac{3y}{-3/2}=1 \;\Rightarrow\; -4x+2y=1 \;\Rightarrow\; y=2x+\tfrac12.$ …

We are told that a point $P(x,y)$ moves so that its distance from $A(0,2)$ is $\frac{1}{\sqrt{2}}$ times its distance from $B(-1,0)$.

This is a ratio condition — it compares two distances. When the ratio is a constant (not equal to 1), the locus is generally a circle (Apollonius circle). The key is to translate the words into algebra, then simplify to the standard circle equation.

1. Write the condition in algebraic form Distance from $P$ to $A$: $\sqrt{(x-0)^2 + (y-2)^2}$ Distance from $P$ to $B$: $\sqrt{(x+1)^2 + (y-0)^2}$ The condition says:

2. Square both sides to remove square roots Squaring is safe because distances are non-negative:

3. Expand both sides

   Left: $x^2 + y^2 - 4y + 4$

   Right: $\frac{1}{2} \left( x^2 + 2x + 1 + y^2 \right) = \frac{1}{2}x^2 + x + \frac{1}{2} + \frac{1}{2}y^2$

   So we have:

4. Clear fractions by multiplying through by 2

5. Bring all terms to one side

Simplify:

6. Complete the square to identify the circle …

The problem asks for the equation of the incircle of a triangle defined by three lines. An incircle is a circle that is tangent to all three sides of a triangle. Its center is called the incenter, and its radius is called the inradius.

The key idea here is to first identify the triangle's vertices and its type. Once we know the type of triangle, especially if it's a right-angled triangle, finding the inradius and incenter becomes straightforward. The equation of a circle with center $(h,k)$ and radius $r$ is given by $(x-h)^2 + (y-k)^2 = r^2$.

1. Identify the vertices of the triangle.

   The three lines forming the triangle are:

   • $L_1: x = 0$ (the y-axis)
   • $L_2: y = 0$ (the x-axis)
   • $L_3: 3x + 4y - 24 = 0$

   To find the vertices, we find the intersection points of these lines:

   • Intersection of $L_1$ and $L_2$: Substitute $x=0$ into $y=0$. This gives the origin, $A(0,0)$.
   • Intersection of $L_1$ and $L_3$: Substitute $x=0$ into $3x + 4y - 24 = 0$. $3(0) + 4y - 24 = 0 \implies 4y = 24 \implies y = 6$. This gives the vertex $B(0,6)$.
   • Intersection of $L_2$ and $L_3$: Substitute $y=0$ into $3x + 4y - 24 = 0$. $3x + 4(0) - 24 = 0 \implies 3x = 24 \implies x = 8$. This gives the vertex $C(8,0)$.

   So, the vertices of the triangle are $A(0,0)$, $B(0,6)$, and $C(8,0)$.

2. Determine the type of triangle and its side lengths.

   Since two sides of the triangle lie along the x-axis ($y=0$) and the y-axis ($x=0$), the triangle has a right angle at the origin $(0,0)$. This is a right-angled triangle.

   Now, let's find the lengths of its sides:

   • Length of side $AC$ (along the x-axis): $a = |8-0| = 8$ units.
   • Length of side $AB$ (along the y-axis): $b = |6-0| = 6$ units.
   • Length of side $BC$ (the hypotenuse): $c = \sqrt{(8-0)^2 + (0-6)^2} = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10$ units.

3. Calculate the inradius ($r$).

   For a right-angled triangle with legs of length $a$ and $b$, and hypotenuse of length $c$, the inradius $r$ can be found using the formula:

   For a right-angled triangle, the inradius $r = \frac{a+b-c}{2}$.

   Using the side lengths we found: $a=8$, $b=6$, $c=10$.

   $r = \frac{8 + 6 - 10}{2} = \frac{14 - 10}{2} = \frac{4}{2} = 2$.

   The inradius is $2$.

4. Determine the coordinates of the incenter. …

The problem asks us to evaluate an Assertion (A) and a Reason (R) related to finding the image of an ellipse in a line. To do this, we need to understand how to find the image of a point in a line, and then how this concept extends to finding the image of an entire curve.

Concept and Intuition

The core idea behind finding the image of a curve (like an ellipse) in a line (the mirror) is to consider every point on the original curve. For each such point, we find its reflection across the given line. The collection of all these reflected points forms the image curve. This is precisely what Reason (R) states.

To implement this, we follow these steps:

1. Take a general point $(x_1, y_1)$ on the original curve.
2. Find the coordinates of its image $(x', y')$ with respect to the line of reflection.
3. Since $(x_1, y_1)$ lies on the original curve, its coordinates satisfy the curve's equation. We then express $x_1$ and $y_1$ in terms of $x'$ and $y'$ using the reflection formulas.
4. Substitute these expressions back into the original curve's equation. The resulting equation, in terms of $x'$ and $y'$, will be the equation of the image curve. Finally, we replace $(x', y')$ with $(x, y)$ to get the standard form of the image curve's equation.

Step-by-Step Derivation

1. Analyze Reason (R):

   Reason (R) states: "The image of a curve ‘C’ in a line L is the locus of the image of every point of C with respect to the line L."

   This statement is the fundamental definition of how reflection transforms a curve. It is conceptually sound and forms the basis for finding the image of any geometric figure.

   Therefore, Reason (R) is true.

2. Analyze Assertion (A):

   The original ellipse is $E_1: \dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$.

   The line of reflection is $L: x + y = 10$.

   We need to find the image of a general point $(x_1, y_1)$ in the line $x+y=10$. Let the image be $(x', y')$.

   The image $(x', y')$ of a point $(x_1, y_1)$ in the line $ax+by+c=0$ is given by:

   $\frac{x'-x_1}{a} = \frac{y'-y_1}{b} = -2 \frac{ax_1+by_1+c}{a^2+b^2}$

   For the line $x+y-10=0$, we have $a=1$, $b=1$, $c=-10$.

   Substituting these values:

   $\frac{x'-x_1}{1} = \frac{y'-y_1}{1} = -2 \frac{x_1+y_1-10}{1^2+1^2}$

   $\frac{x'-x_1}{1} = \frac{y'-y_1}{1} = -(x_1+y_1-10)$

   From the first equality:

   $x'-x_1 = -(x_1+y_1-10)$

   $x' = x_1 - x_1 - y_1 + 10$

   $x' = 10 - y_1$

   From the second equality:

   $y'-y_1 = -(x_1+y_1-10)$

   $y' = y_1 - x_1 - y_1 + 10$

   $y' = 10 - x_1$

   So, the image of a point $(x_1, y_1)$ in the line $x+y=10$ is $(10-y_1, 10-x_1)$.

3. Find the equation of the image curve: …

The point lies on the parabola, so it fixes $a$. Substituting the endpoint into $y^2=\dfrac{8}{a}\,|x|$ gives $16=\dfrac{8}{a}$, hence $a=\tfrac12$ and the parabola is $y^2 = 16x$ (magnitude of latus rectum $4A=16$, so $A=4$).

Focus is at distance $A=4$ from the vertex and the directrix is $|x|=4$.

Endpoint $1$ at $x=1$: focal distance $= x + A = 1 + 4 = 5$. …

The core idea here is to use the equation of a conic section to determine the position of a point relative to it. For a general conic section given by the equation $S(x, y) = Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$, if we substitute the coordinates $(x_1, y_1)$ of a point into the expression $S(x, y)$, the sign of the resulting value $S(x_1, y_1)$ tells us whether the point is inside, on, or outside the curve.

Specifically, for a circle or an ellipse:

• If $S(x_1, y_1) < 0$, the point $(x_1, y_1)$ lies inside the conic.
• If $S(x_1, y_1) = 0$, the point $(x_1, y_1)$ lies on the conic.
• If $S(x_1, y_1) > 0$, the point $(x_1, y_1)$ lies outside the conic.

This works because the equation $S(x, y) = 0$ defines the boundary. When you move away from the center of an ellipse or circle, the value of $S(x, y)$ increases (or decreases, depending on the sign convention chosen for $S(x,y)$). The standard convention for $S(x,y)$ is to have the constant term be negative when the curve passes through the origin, or to have the right-hand side be 1 for standard forms. In our case, $x^2+y^2-r^2=0$ for a circle and $\frac{x^2}{a^2}+\frac{y^2}{b^2}-1=0$ for an ellipse, the "inside" region corresponds to $S(x,y) < 0$.

Let's apply this concept to the given ellipse and circle, and the points A and B.

1. Standardize the equations:

   The given ellipse (E) is $4x^2 + 9y^2 - 36 = 0$. We define $S_E(x, y) = 4x^2 + 9y^2 - 36$.

   The given circle (C) is $x^2 + y^2 - 9 = 0$. We define $S_C(x, y) = x^2 + y^2 - 9$.

2. Check point A(1, 2) with Circle (C):

   Substitute $x=1$ and $y=2$ into $S_C(x, y)$:

   $S_C(1, 2) = (1)^2 + (2)^2 - 9 = 1 + 4 - 9 = 5 - 9 = -4$.

   Since $S_C(1, 2) = -4 < 0$, point A lies inside the circle (C).

3. Check point A(1, 2) with Ellipse (E):

   Substitute $x=1$ and $y=2$ into $S_E(x, y)$:

   $S_E(1, 2) = 4(1)^2 + 9(2)^2 - 36 = 4(1) + 9(4) - 36 = 4 + 36 - 36 = 4$.

   Since $S_E(1, 2) = 4 > 0$, point A lies outside the ellipse (E). …

The key is to translate the geometric conditions into algebra. A circle passes through the origin, so its equation has no constant term. It also cuts a chord of fixed length on a vertical line — that gives a relation between the centre’s coordinates.

Let the centre be $C(h, k)$ and the radius be $r$. Since the circle passes through $(0,0)$, we have

Now the line $x = 1$ cuts the circle. The chord on this line has length 2. The distance from the centre to the line $x = 1$ is $|h - 1|$. The half-length of the chord is $\sqrt{r^2 - (\text{distance})^2}$. So the full chord length is

Divide by 2:

Square both sides:

Now substitute $r^2 = h^2 + k^2$:

Expand $(h-1)^2 = h^2 - 2h + 1$:

Simplify:

So

Thus

Replace $(h, k)$ with $(x, y)$ for the locus:

This is $y^2 = -2(x - 1)$, which is a leftward-opening parabola with vertex at $(1, 0)$.