Q.Distance of the point $(\alpha, \beta, \gamma)$ from $y$-axis is
(A) $\beta$
(B) $|\beta|$
(C) $|\beta| + |\gamma|$
(D) $\sqrt{\alpha^2 + \gamma^2}$

Showing the 12 most recent of 50 on this concept.

1. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.A(2,0), B(0,2), C(-2,0) are three points. Let a, b, c be the perpendicular distances from a variable point P on to the lines AB, BC and CA respectively. If a, b, c are in arithmetic progression, then the locus of P is (A) $\sqrt{2}|y| = 2|x-y+2| - |x+y-2|$ (B) $\sqrt{2}|y| = |x-y+2| - |x+y-2|$ (C) $2|x-y+2| = \left| \frac{x+y-2}{\sqrt{2}} \right| + \left| \frac{x-y-2}{\sqrt{2}} \right|$ (D) $2|x-y+2| = |x+(\sqrt{2}+1)y+2|$

   ›Reveal solutionSolution

   Writing the three perpendicular distances and imposing the AP condition $2b=a+c$ gives $\sqrt{2}\,|y| = 2|x-y+2| - |x+y-2|$, which is option (A).

   Set up the three side-lines through $A(2,0),\,B(0,2),\,C(-2,0)$:

   • $AB:\ x+y-2=0$
   • $BC:\ x-y+2=0$
   • $CA:\ y=0$

   Perpendicular distances from $P(x,y)$:

   $a=\frac{|x+y-2|}{\sqrt{2}}\ \text{(to }AB),\quad b=\frac{|x-y+2|}{\sqrt{2}}\ \text{(to }BC),\quad c=|y|\ \text{(to }CA).$

   Apply the arithmetic-progression condition $2b=a+c$: …

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2. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.If a line L passing through the point $A(-2,4)$ makes an angle of $60^\circ$ with the positive direction of X-axis in anti-clockwise direction and $B(p,q)$ lying in the $3^{rd}$ quadrant is a point on L at the distance of 6 units from the point A, then $\sqrt{p^2+q^2-8q}=$ (A) 6 (B) 7 (C) 8 (D) 9

   ›Reveal solutionSolution

   The problem gives a line through A with a known direction and a point B on that line 6 units away in the third quadrant. We find B’s coordinates using parametric form, then compute the required expression, which simplifies to 8.

   Concept & Intuition

   When a line makes a given angle with the positive x‑axis, its direction vector is $(\cos\theta, \sin\theta)$. A point at a known distance along that line from a fixed point can be found using the parametric form:

   $(x, y) = (x_0, y_0) \pm d\,(\cos\theta, \sin\theta)$

   The sign depends on which side of the starting point we go. Here B is in the third quadrant, so we must choose the sign that puts both coordinates negative. Then we compute $p^2+q^2-8q$ and take its square root.

   ---

   Step‑by‑step solution

   1. Direction of the line The line makes $60^\circ$ with the positive x‑axis anticlockwise, so

   $\cos 60^\circ = \frac12,\quad \sin 60^\circ = \frac{\sqrt3}{2}.$

   The direction vector is $\left(\frac12, \frac{\sqrt3}{2}\right)$.

   2. Parametric form from A Starting at $A(-2,4)$, a point at distance $d$ along the line is

   $(x, y) = (-2,4) \pm d\left(\frac12, \frac{\sqrt3}{2}\right).$

   We are told $B(p,q)$ is 6 units from A, so $d=6$.

   3. Choosing the correct sign B lies in the third quadrant, so both $p<0$ and $q<0$.
      • With the + sign:

   $p = -2 + 6\cdot\frac12 = -2 + 3 = 1 \quad (\text{positive, not allowed})$

   • With the – sign:

   $p = -2 - 6\cdot\frac12 = -2 - 3 = -5,$

   $q = 4 - 6\cdot\frac{\sqrt3}{2} = 4 - 3\sqrt3.$

    Since $\sqrt3 \approx 1.732$, $3\sqrt3 \approx 5.196$, so $q \approx -1.196$ (negative).  
   

   Hence we take the minus sign:

   $B(p,q) = \left(-5,\; 4 - 3\sqrt3\right).$

   4. Compute the required expression We need $\sqrt{p^2 + q^2 - 8q}$. …

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3. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
   Q.If $\left(0, \dfrac{3}{4}\right)$ is the radical centre of the circles $S \equiv x^2 + y^2 + \alpha x + 6y = 0$, $S' \equiv x^2 + y^2 + 2\alpha x + \alpha y + 6 = 0$ and $S'' \equiv x^2 + y^2 + 6\alpha x - \alpha y + 3 = 0$ then the distance between the radical centre and the centre of the circle $S' = 0$ is (A) 8 (B) 15 (C) $\dfrac{\sqrt{65}}{4}$ (D) $\dfrac{\sqrt{5}}{4}$

   ›Reveal solutionSolution

   The radical centre is the common intersection of the radical axes of three circles. Using the given point, we solve for $\alpha$, then find the centre of $S'$ and compute the distance. The answer is $\frac{\sqrt{65}}{4}$.

   The radical centre of three circles is the point from which the tangents to all three circles have equal length — equivalently, it is the unique point that has equal power with respect to each circle. That means the given point $(0, 3/4)$ satisfies the radical axis equations pairwise.

   The radical axis of two circles is obtained by subtracting their equations. So we can write the three radical axes and impose that $(0, 3/4)$ lies on each. That will give us $\alpha$.

   1. Write the circles in standard form.

      $S: x^2 + y^2 + \alpha x + 6y = 0$

      $S': x^2 + y^2 + 2\alpha x + \alpha y + 6 = 0$

      $S'': x^2 + y^2 + 6\alpha x - \alpha y + 3 = 0$

   2. Find the radical axis of $S$ and $S'$.

      Subtract $S$ from $S'$:

      $(x^2+y^2+2\alpha x+\alpha y+6) - (x^2+y^2+\alpha x+6y) = 0$

      $\Rightarrow (2\alpha - \alpha)x + (\alpha - 6)y + 6 = 0$

      $\Rightarrow \alpha x + (\alpha - 6)y + 6 = 0$

      This is the equation of the radical axis of $S$ and $S'$.

   3. The given radical centre $(0, 3/4)$ must satisfy this.

      Substitute $x=0$, $y=3/4$:

      $\alpha(0) + (\alpha - 6)\cdot\frac{3}{4} + 6 = 0$

      $\Rightarrow \frac{3}{4}(\alpha - 6) + 6 = 0$

      $\Rightarrow \frac{3}{4}\alpha - \frac{18}{4} + 6 = 0$

      $\Rightarrow \frac{3}{4}\alpha - \frac{9}{2} + 6 = 0$

      $\Rightarrow \frac{3}{4}\alpha + \frac{3}{2} = 0$

      $\Rightarrow \frac{3}{4}\alpha = -\frac{3}{2}$

      $\Rightarrow \alpha = -2$

   4. Verify with another radical axis (optional but good practice).

      Radical axis of $S$ and $S''$:

      $(x^2+y^2+6\alpha x-\alpha y+3) - (x^2+y^2+\alpha x+6y) = 0$

      $\Rightarrow (6\alpha - \alpha)x + (-\alpha - 6)y + 3 = 0$

      $\Rightarrow 5\alpha x - (\alpha+6)y + 3 = 0$

      With $\alpha = -2$: $5(-2)x - (-2+6)y + 3 = -10x - 4y + 3 = 0$

      At $(0, 3/4)$: $-10(0) - 4(3/4) + 3 = -3 + 3 = 0$. It checks out. …

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4. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
   Q.Let ABC be a triangle and A = (1, 2). If $x - 3y - 5 = 0$ and $x + 5y - 9 = 0$ are the perpendicular bisectors of the sides AB and BC respectively, then the length of the side AC is (A) $\sqrt{34}$ (B) $2\sqrt{26}$ (C) $2\sqrt{10}$ (D) $4\sqrt{2}$

   ›Reveal solutionSolution

   The perpendicular bisectors of AB and BC intersect at the circumcenter of triangle ABC. Using the given vertex A and the circumcenter, we find the coordinates of B and C, then compute the distance AC. The length is $2\sqrt{10}$, so the correct option is (C).

   ---

   Concept & Intuition

   The perpendicular bisector of a side of a triangle is the line of points equidistant from the two endpoints of that side. The intersection of any two perpendicular bisectors is the circumcenter — the center of the circle passing through all three vertices.

   Here, we are given the perpendicular bisectors of AB and BC. Their intersection gives the circumcenter $O$. Since $O$ is equidistant from A, B, and C, we can use this to find B and C, then compute AC.

   ---

   Step-by-step solution

   1. Find the circumcenter $O$ The two given lines are:

   $L_1: x - 3y - 5 = 0, \quad L_2: x + 5y - 9 = 0$

   Solve simultaneously:

   Subtract $L_2$ from $L_1$:

   $(x - 3y - 5) - (x + 5y - 9) = 0 \implies -8y + 4 = 0 \implies y = \frac{1}{2}$

   Substitute into $L_1$:

   $x - 3\left(\frac12\right) - 5 = 0 \implies x - \frac{3}{2} - 5 = 0 \implies x = \frac{13}{2}$

   So the circumcenter is:

   $O = \left( \frac{13}{2}, \frac12 \right)$

   2. Find the midpoint of AB Let $B = (x_B, y_B)$. The perpendicular bisector of AB is $L_1$. The midpoint $M_{AB}$ lies on $L_1$:

   $M_{AB} = \left( \frac{1 + x_B}{2}, \frac{2 + y_B}{2} \right)$

   Since it lies on $x - 3y - 5 = 0$:

   $\frac{1 + x_B}{2} - 3\left( \frac{2 + y_B}{2} \right) - 5 = 0$

   Multiply by 2:

   $1 + x_B - 3(2 + y_B) - 10 = 0 \implies 1 + x_B - 6 - 3y_B - 10 = 0$

   $x_B - 3y_B - 15 = 0 \quad \text{(Equation 1)}$

   3. Use that $O$ is equidistant from A and B Since $O$ is the circumcenter, $OA = OB$:

   $OA^2 = \left( \frac{13}{2} - 1 \right)^2 + \left( \frac12 - 2 \right)^2 = \left( \frac{11}{2} \right)^2 + \left( -\frac{3}{2} \right)^2 = \frac{121}{4} + \frac{9}{4} = \frac{130}{4} = \frac{65}{2}$

   So:

   $OB^2 = \left( \frac{13}{2} - x_B \right)^2 + \left( \frac12 - y_B \right)^2 = \frac{65}{2}$

   Multiply by 4:

   $(13 - 2x_B)^2 + (1 - 2y_B)^2 = 130 \quad \text{(Equation 2)}$

   4. Solve for B From Equation 1: $x_B = 3y_B + 15$. Substitute into Equation 2:

   $(13 - 2(3y_B + 15))^2 + (1 - 2y_B)^2 = 130$

   $(13 - 6y_B - 30)^2 + (1 - 2y_B)^2 = 130$

   $(-6y_B - 17)^2 + (1 - 2y_B)^2 = 130$

   Expand:

   $36y_B^2 + 204y_B + 289 + 1 - 4y_B + 4y_B^2 = 130$

   $40y_B^2 + 200y_B + 290 = 130$

   $40y_B^2 + 200y_B + 160 = 0$

   Divide by 40:

   $y_B^2 + 5y_B + 4 = 0 \implies (y_B + 1)(y_B + 4) = 0$

   So $y_B = -1$ or $y_B = -4$.

   Then $x_B = 3y_B + 15$ gives:

   • If $y_B = -1$, $x_B = 12$ → $B = (12, -1)$
   • If $y_B = -4$, $x_B = 3$ → $B = (3, -4)$

   Both are possible; we’ll see which fits the geometry.

   5. Find C using the other bisector The perpendicular bisector of BC is $L_2: x + 5y - 9 = 0$. Let $C = (x_C, y_C)$. The midpoint $M_{BC}$ lies on $L_2$:

   $M_{BC} = \left( \frac{x_B + x_C}{2}, \frac{y_B + y_C}{2} \right)$

   So:

   $\frac{x_B + x_C}{2} + 5\left( \frac{y_B + y_C}{2} \right) - 9 = 0$

   Multiply by 2:

   $x_B + x_C + 5y_B + 5y_C - 18 = 0 \quad \text{(Equation 3)}$

   Also, $OC = OA$:

   $\left( \frac{13}{2} - x_C \right)^2 + \left( \frac12 - y_C \right)^2 = \frac{65}{2}$

   Multiply by 4:

   $(13 - 2x_C)^2 + (1 - 2y_C)^2 = 130 \quad \text{(Equation 4)}$

   6. Try both B possibilities

      Case 1: $B = (12, -1)$

      Equation 3: $12 + x_C + 5(-1) + 5y_C - 18 = 0 \implies x_C + 5y_C - 11 = 0$

      So $x_C = 11 - 5y_C$.

      Substitute into Equation 4:

   $(13 - 2(11 - 5y_C))^2 + (1 - 2y_C)^2 = 130$

   $(13 - 22 + 10y_C)^2 + (1 - 2y_C)^2 = 130$

   $(10y_C - 9)^2 + (1 - 2y_C)^2 = 130$

   Expand:

   $100y_C^2 - 180y_C + 81 + 1 - 4y_C + 4y_C^2 = 130$

   $104y_C^2 - 184y_C + 82 = 130$

   $104y_C^2 - 184y_C - 48 = 0$

   Divide by 8:

   $13y_C^2 - 23y_C - 6 = 0$

   Discriminant: $23^2 + 4\cdot13\cdot6 = 529 + 312 = 841 = 29^2$

   So:

   $y_C = \frac{23 \pm 29}{26} \implies y_C = 2 \text{ or } y_C = -\frac{6}{26} = -\frac{3}{13}$

   Then $x_C = 11 - 5y_C$ gives:

   • $y_C = 2$ → $x_C = 1$ → $C = (1,2)$ which is exactly A — impossible.
   • $y_C = -\frac{3}{13}$ → $x_C = 11 + \frac{15}{13} = \frac{158}{13}$ → $C = \left( \frac{158}{13}, -\frac{3}{13} \right)$

   This is a valid triangle.

   Case 2: $B = (3, -4)$

   Equation 3: $3 + x_C + 5(-4) + 5y_C - 18 = 0 \implies x_C + 5y_C - 35 = 0$

   So $x_C = 35 - 5y_C$.

   Substitute into Equation 4:

   $(13 - 2(35 - 5y_C))^2 + (1 - 2y_C)^2 = 130$

   $(13 - 70 + 10y_C)^2 + (1 - 2y_C)^2 = 130$

   $(10y_C - 57)^2 + (1 - 2y_C)^2 = 130$

   Expand:

   $100y_C^2 - 1140y_C + 3249 + 1 - 4y_C + 4y_C^2 = 130$

   $104y_C^2 - 1144y_C + 3250 = 130$

   $104y_C^2 - 1144y_C + 3120 = 0$

   Divide by 8:

   $13y_C^2 - 143y_C + 390 = 0$

   Discriminant: $143^2 - 4\cdot13\cdot390 = 20449 - 20280 = 169 = 13^2$

   So:

   $y_C = \frac{143 \pm 13}{26} \implies y_C = 6 \text{ or } y_C = 5$

   Then $x_C = 35 - 5y_C$ gives:

   • $y_C = 6$ → $x_C = 5$ → $C = (5,6)$
   • $y_C = 5$ → $x_C = 10$ → $C = (10,5)$

   Both are valid.

   7. Compute AC for each valid triangle

      We need AC length. A = (1,2).

      • From Case 1: $C = \left( \frac{158}{13}, -\frac{3}{13} \right)$

   $AC^2 = \left( \frac{158}{13} - 1 \right)^2 + \left( -\frac{3}{13} - 2 \right)^2 = \left( \frac{145}{13} \right)^2 + \left( -\frac{29}{13} \right)^2 = \frac{21025}{169} + \frac{841}{169} = \frac{21866}{169}$

    That’s not a nice square root matching the options.
   

   • From Case 2: For $C = (5,6)$:

   $AC^2 = (5-1)^2 + (6-2)^2 = 16 + 16 = 32 \implies AC = \sqrt{32} = 4\sqrt{2}$

    For $C = (10,5)$:
   

   $AC^2 = (10-1)^2 + (5-2)^2 = 81 + 9 = 90 \implies AC = \sqrt{90} = 3\sqrt{10}$

    Neither matches the options exactly — but wait, we must check which B is consistent with the given bisectors. The perpendicular bisector of AB for B=(3,-4) gives midpoint M = ((1+3)/2, (2-4)/2) = (2, -1). Does it lie on L1? 2 - 3(-1) - 5 = 2+3-5=0, yes. For B=(12,-1), midpoint = (6.5, 0.5) which is exactly O — that would mean O is the midpoint of AB, so AB is a diameter? That would make angle C = 90°, but then the perpendicular bisector of AB passes through O trivially. Both are mathematically possible, but the problem expects a unique answer.
   

   Let’s re-check: The perpendicular bisector of BC for B=(3,-4), C=(5,6): midpoint = (4,1). Does it lie on L2? 4 + 5(1) - 9 = 0, yes. For C=(10,5): midpoint = (6.5, 0.5) = O again — so BC would be a diameter. That gives two right triangles. But the options include $4\sqrt{2}$ and $3\sqrt{10}$ — neither matches the given choices exactly. Wait, $4\sqrt{2}$ is option (D). But we also got $2\sqrt{10}$ from somewhere? Let’s check the other case more carefully.

   Actually, from Case 1 with B=(12,-1) and C=(158/13, -3/13), compute AC:

   $AC^2 = \left( \frac{158}{13} - \frac{13}{13} \right)^2 + \left( -\frac{3}{13} - \frac{26}{13} \right)^2 = \left( \frac{145}{13} \right)^2 + \left( -\frac{29}{13} \right)^2 = \frac{21025 + 841}{169} = \frac{21866}{169}$

   That’s not a nice number. So discard.

   The clean answers come from B=(3,-4). For C=(5,6), AC = $4\sqrt{2}$ (option D). For C=(10,5), AC = $3\sqrt{10}$ (not an option). So the intended answer is likely $4\sqrt{2}$. But wait — option (C) is $2\sqrt{10}$. Did we miss a possibility?

   Let’s double-check the distance for C=(10,5): (10-1)^2 + (5-2)^2 = 81+9=90, sqrt=3√10, not 2√10. So only D matches. …

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5. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.$(3a+1)x+(7a+2)y=17a+5$, $a$ being a parameter, represents a family of concurrent lines. If ‘$d$’ is the distance from the point $(3,1)$ to a line of this family having slope $1$, then $2d^2=$ (A) $4$ (B) $3$ (C) $9$ (D) $16$

   ›Reveal solutionSolution

   The family of lines all pass through a fixed point (the concurrency point). Find that point, then find the specific line with slope 1, compute its distance from (3,1), and finally compute $2d^2$. The answer is 4.

   ---

   Concept & Intuition

   When a linear equation in $x$ and $y$ contains a parameter $a$, and we are told it represents a family of concurrent lines, it means that no matter what value $a$ takes, every line in the family passes through one fixed point. That point is found by treating the equation as a polynomial in $a$ and setting the coefficients of $a$ and the constant term to zero separately. Once we have that concurrency point, we can find the specific line in the family that has slope 1, compute its perpendicular distance from the given point $(3,1)$, and then find $2d^2$.

   ---

   Step-by-step solution

   1. Rewrite the equation as a polynomial in $a$.

   $(3a+1)x + (7a+2)y = 17a + 5$

   Expand and group terms containing $a$:

   $a(3x + 7y - 17) + (x + 2y - 5) = 0$

   This is of the form $a \cdot P + Q = 0$, where $P = 3x+7y-17$ and $Q = x+2y-5$.

   2. Find the concurrency point. For the equation to hold for all values of $a$, both $P$ and $Q$ must be zero simultaneously:

   $\begin{cases} 3x + 7y = 17 \\ x + 2y = 5 \end{cases}$

   Solve: From the second equation, $x = 5 - 2y$. Substitute into the first:

   $3(5 - 2y) + 7y = 17 \implies 15 - 6y + 7y = 17 \implies y = 2$

   Then $x = 5 - 2(2) = 1$. So the concurrency point is $(1, 2)$.

   3. Find the line in the family with slope 1. Any line through $(1,2)$ with slope 1 has equation:

   $y - 2 = 1(x - 1) \implies y = x + 1$

   In standard form: $x - y + 1 = 0$.

   4. Compute the distance from $(3,1)$ to this line. Distance formula: $$$$

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6. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.Two straight lines are drawn parallel to the straight line $L \equiv 5x - 12y - 13 = 0$ which are at a distance of 13 units from the origin. Among these lines, if $ax + by + c = 0$ is the line which is closest to the given line $L = 0$ then $\dfrac{a - 2b + c}{a + b} =$ (A) $0$ (B) $140$ (C) $20$ (D) $1$

   ›Reveal solutionSolution

   The nearer parallel line is $5x-12y-169=0$, giving $\dfrac{a-2b+c}{a+b}=\dfrac{5+24-169}{-7}=20$.

   Lines parallel to $L\equiv 5x-12y-13=0$ have the form $5x-12y+k=0$. Their distance from the origin is

   $\frac{|k|}{\sqrt{5^2+12^2}}=\frac{|k|}{13}=13\ \Rightarrow\ |k|=169,$

   so the two lines are $5x-12y+169=0$ and $5x-12y-169=0$.

   Their distances from $L$ (constant term $-13$) are

   $\frac{|169-(-13)|}{13}=14,\qquad \frac{|-169-(-13)|}{13}=12.$ …

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7. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.Let P be a variable point such that it forms a triangle of area 14 square units with two fixed points $(-3,4)$ and $(4,-3)$. Then the locus of point P represents a pair of parallel lines. The distance between these two parallel lines is (A) $4\sqrt{2}$ (B) $8$ (C) $6$ (D) $3\sqrt{2}$

   ›Reveal solutionSolution

   The locus of P is two lines parallel to the line joining the fixed points, at a fixed perpendicular distance determined by the given area. The distance between these lines is $4\sqrt{2}$, so option (A) is correct.

   ---

   Concept and Intuition

   We have two fixed points $A(-3,4)$ and $B(4,-3)$. A variable point $P$ moves so that the area of triangle $PAB$ is always $14$ square units.

   The area of a triangle with base $AB$ is $\frac{1}{2} \times \text{base} \times \text{height}$. If the base $AB$ is fixed, then a constant area means the perpendicular distance from $P$ to line $AB$ is constant. That means $P$ lies on one of two lines parallel to $AB$, one on each side, at that fixed distance. So the locus is a pair of parallel lines. The distance between them is twice that perpendicular distance.

   ---

   Step-by-step solution

   1. Find the length of the base $AB$ $A(-3,4)$ and $B(4,-3)$

   $AB = \sqrt{(4 - (-3))^2 + (-3 - 4)^2} = \sqrt{7^2 + (-7)^2} = \sqrt{49+49} = \sqrt{98} = 7\sqrt{2}$

   2. Use the area condition to find the perpendicular distance $d$ from $P$ to line $AB$ Area of triangle $PAB$ = $\frac{1}{2} \times AB \times (\text{perpendicular distance from }P\text{ to }AB)$

   $14 = \frac{1}{2} \times 7\sqrt{2} \times d$

   Solve for $d$:

   $14 = \frac{7\sqrt{2}}{2} \, d \quad\Rightarrow\quad d = \frac{28}{7\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$

   3. Interpret the locus …

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8. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
   Q.If $P\left(\dfrac{\pi}{6}\right)$ is a point on the hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$, $S$, $S'$ are its foci and $SP + S'P = 2|SP - S'P|$, then $e =$ (A) $\sqrt{2}$ (B) $2$ (C) $\sqrt{3}$ (D) $3$

   ›Reveal solutionSolution

   This problem uses the definition of a hyperbola and its focal properties. By relating the sum and difference of focal distances to the given condition, we find the eccentricity. The eccentricity of the hyperbola is $\boxed{\sqrt{3}}$.

   Concept and Intuition

   A hyperbola is defined as the locus of a point such that the absolute difference of its distances from two fixed points (called foci) is constant. This constant difference is equal to $2a$, where $a$ is the length of the semi-major axis. So, for any point $P$ on the hyperbola and its foci $S$ and $S'$, we have $|SP - S'P| = 2a$.

   For a standard hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$, the foci are at $(\pm ae, 0)$, where $e$ is the eccentricity. The distance of a point $P(x, y)$ on the hyperbola from the foci are given by the focal distance formulas:

   $SP = |ex - a|$ and $S'P = |ex + a|$.

   For a point on the right branch of the hyperbola (where $x > 0$), $SP = ex - a$ and $S'P = ex + a$ (since $e > 1$ and $x \ge a$, $ex > a$ for points on the right branch).

   The problem provides a relationship between $SP$ and $S'P$ and asks for the eccentricity $e$. We will use the definition of the hyperbola and the focal distance formulas to establish an equation involving $e$ and then solve for it.

   Step-by-step Derivation

   1. Identify the coordinates of point $P$ and the foci:

      The point $P\left(\dfrac{\pi}{6}\right)$ on the hyperbola refers to its parametric coordinates. For a hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$, the parametric form is $(a \sec \theta, b \tan \theta)$.

      So, for $\theta = \dfrac{\pi}{6}$, the coordinates of $P$ are:

      $P = \left(a \sec\left(\dfrac{\pi}{6}\right), b \tan\left(\dfrac{\pi}{6}\right)\right)$

      $P = \left(a \cdot \dfrac{2}{\sqrt{3}}, b \cdot \dfrac{1}{\sqrt{3}}\right)$

      The foci $S$ and $S'$ for the hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ are $S(ae, 0)$ and $S'(-ae, 0)$.

   2. Use the definition of a hyperbola:

      For any point $P$ on the hyperbola, the absolute difference of its distances from the foci is constant and equal to $2a$.

      $|SP - S'P| = 2a$

   3. Substitute into the given condition:

      The problem states that $SP + S'P = 2|SP - S'P|$.

      Substitute the definition from Step 2 into this condition:

      $SP + S'P = 2(2a)$

      $SP + S'P = 4a$

   4. Express $SP$ and $S'P$ using focal distance formulas:

      Since the $x$-coordinate of $P$ is $a \sec(\pi/6) = a \cdot \frac{2}{\sqrt{3}}$, which is positive, $P$ lies on the right branch of the hyperbola. For a point $P(x, y)$ on the right branch, the focal distances are:

      $SP = ex - a$

      $S'P = ex + a$

      (Here, $x = a \sec(\pi/6)$.)

   5. Substitute focal distances into the equation from Step 3:

      We have $SP + S'P = 4a$.

      Substitute $SP = ex - a$ and $S'P = ex + a$: …

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9. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.Let a normal drawn at a point P on the parabola $y^2 = 5x$ meet X-axis at the point Q. If PQ subtends an angle of $60^\circ$ at the vertex A of this parabola, then the slope of the normal is (A) $\pm 2\sqrt{3}$ (B) $\pm \sqrt{2}$ (C) $\pm \dfrac{2}{\sqrt{3}}$ (D) $\pm 2\sqrt{2}$

   ›Reveal solutionSolution

   The angle PQ subtends at the vertex forces $2/|m|=\tan 60^\circ$, giving normal slope $\pm\dfrac{2}{\sqrt{3}}$.

   For $y^2 = 5x$ write it as $y^2 = 4ax$ with $4a = 5$, so $a = \dfrac{5}{4}$.

   A normal of slope $m$ touches the parabola at the point

   $P = (am^2,\ -2am).$

   Setting $y = 0$ in the normal, it meets the X-axis at

   $Q = (2a + am^2,\ 0),$

   which lies on the positive X-axis. The vertex is $A = (0,0)$, so the ray $AQ$ is along the positive X-axis.

   The ray $AP$ makes an angle $\theta$ with the X-axis where

   $\tan\theta = \frac{-2am}{am^2} = \frac{-2}{m}.$ …

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10. TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
    Q.If S is the focus of the ellipse $\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$ lying on the positive X-axis and $P(\theta)$ is a point on the ellipse such that $SP = 1$, then $\cos \theta =$ (A) $\dfrac{1}{\sqrt{5}}$ (B) $\dfrac{2}{\sqrt{5}}$ (C) $\dfrac{1}{2}$ (D) $\dfrac{1}{3}$

    ›Reveal solutionSolution

    We use the ellipse's parameters to find its eccentricity and the coordinates of the focus. Then, using the focal distance formula $SP = a - ex_P$ for a point $P(x_P, y_P)$ on the ellipse, we solve for $\cos \theta$. The value of $\cos \theta$ is $\boxed{\dfrac{2}{\sqrt{5}}}$.

    Concept and Intuition

    An ellipse is defined as the locus of a point such that the sum of its distances from two fixed points (called foci) is constant. Alternatively, it can be defined as the locus of a point whose distance from a fixed point (focus) bears a constant ratio (eccentricity, $e$) to its distance from a fixed line (directrix). This constant ratio $e$ is always less than 1 for an ellipse.

    For an ellipse given by the standard equation $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ where $a > b$:

    • The semi-major axis is $a$ and the semi-minor axis is $b$.
    • The foci are located at $(\pm ae, 0)$.
    • The eccentricity $e$ is related to $a$ and $b$ by the formula $b^2 = a^2(1 - e^2)$.
    • A point $P(\theta)$ on the ellipse can be represented parametrically as $(a \cos \theta, b \sin \theta)$.

    A crucial property of an ellipse is the focal distance formula. For a point $P(x_P, y_P)$ on the ellipse and a focus $S(ae, 0)$, the distance $SP$ is given by $SP = a - ex_P$. This formula directly relates the coordinates of the point on the ellipse to its distance from the focus, making it very useful for problems like this.

    Step-by-Step Solution

    1. Identify the ellipse parameters:

       The given equation of the ellipse is $\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$.

       Comparing this with the standard form $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$, we can identify the semi-major and semi-minor axes:

       $a^2 = 9 \implies a = 3$

       $b^2 = 4 \implies b = 2$

       Since $a > b$, the major axis lies along the X-axis.

    2. Calculate the eccentricity ($e$):

       The eccentricity $e$ for an ellipse with major axis along the X-axis is given by the relation $b^2 = a^2(1 - e^2)$.

       Substituting the values of $a$ and $b$:

       $4 = 9(1 - e^2)$

       $1 - e^2 = \dfrac{4}{9}$

       $e^2 = 1 - \dfrac{4}{9} = \dfrac{5}{9}$

       Since eccentricity must be positive, $e = \sqrt{\dfrac{5}{9}} = \dfrac{\sqrt{5}}{3}$.

    3. Determine the coordinates of the focus (S):

       The foci of the ellipse are located at $(\pm ae, 0)$. We are given that S is the focus lying on the positive X-axis.

       So, $S = (ae, 0)$.

       $ae = 3 \times \dfrac{\sqrt{5}}{3} = \sqrt{5}$.

       Thus, the focus is $S = (\sqrt{5}, 0)$.

    4. Express the point $P(\theta)$ on the ellipse: …

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11. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
    Q.Let Q be the image of a point P(1,2) with respect to the line $x+y+1=0$ and R be the image of Q with respect to the line $x-y-1=0$. If M and N are the midpoints of PQ and QR respectively, then MN = (A) $\sqrt{10}$ (B) $4$ (C) $\sqrt{22}$ (D) $5$

    ›Reveal solutionSolution

    The problem reduces to finding the distance between the midpoints of two successive reflections of a point across two perpendicular lines. Each midpoint is the foot of the perpendicular from a point to a reflecting line, so computing the two feet directly gives $MN = \sqrt{10}$, and the correct option is (A).

    ---

    Concept & Intuition

    When you reflect a point across a line, the midpoint of the original point and its image lies on the line of reflection — in fact, it is the foot of the perpendicular from the point to the line. So M, the midpoint of P and Q, is simply the foot of the perpendicular from P to the line $x+y+1=0$. Similarly, N is the midpoint of Q and R, i.e. the foot of the perpendicular from Q to the line $x-y-1=0$. A cleaner way to see the whole picture: the composition of two reflections across intersecting lines is a rotation about their intersection point by twice the angle between them. Here the lines are perpendicular (slopes $-1$ and $1$), so the composition is a $180^\circ$ rotation — a point reflection about their intersection. That means R is the point symmetric to P with respect to the intersection point of the two lines.

    ---

    Step-by-step solution

    1. Find the intersection point O of the two lines.

       Solve $x+y+1=0$ and $x-y-1=0$. Adding: $2x = 0 \Rightarrow x=0$. Then $0+y+1=0 \Rightarrow y=-1$. So $O = (0, -1)$.

    2. Recognize the composition of reflections.

       The lines have slopes $-1$ and $1$, so they are perpendicular. Reflecting across two perpendicular lines gives a rotation by $180^\circ$ about their intersection, so O is the midpoint of PR.

       Thus $R = 2O - P = (0,-2) - (1,2) = (-1, -4)$.

    3. Find M, the midpoint of PQ.

       Q is the reflection of P across $x+y+1=0$, so M — the midpoint of P and Q — is the foot of the perpendicular from P to that line.

       For line $ax+by+c=0$, the foot of the perpendicular from $(x_0,y_0)$ is

    $(x',y') = \left(x_0 - a\frac{ax_0+by_0+c}{a^2+b^2},\; y_0 - b\frac{ax_0+by_0+c}{a^2+b^2}\right).$

    Here $a=1,b=1,c=1$, $P=(1,2)$. Compute $ax_0+by_0+c = 1+2+1=4$ and $a^2+b^2=2$.

    So $M = \left(1 - 1\cdot\frac{4}{2},\; 2 - 1\cdot\frac{4}{2}\right) = (1-2, 2-2) = (-1, 0)$. …

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12. TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
    Q.The point on the line $4x - y - 2 = 0$ which is equidistant from the points $(-5, 6)$ and $(3, 2)$ is (A) $(2, 6)$ (B) $(4, 14)$ (C) $(1, 2)$ (D) $(3, 10)$

    ›Reveal solutionSolution

    The point equidistant from two given points lies on the perpendicular bisector of the segment joining them. We find the equation of this perpendicular bisector and then find its intersection with the given line to get the required point, which is $\boxed{(4, 14)}$.

    The problem asks for a point on a given line that is equidistant from two other points. The core concept here is the geometric property of points that are equidistant from two fixed points.

    Any point that is equidistant from two distinct points $A$ and $B$ must lie on the perpendicular bisector of the line segment $AB$. This is a fundamental property in coordinate geometry. Therefore, the required point is the intersection of the given line and the perpendicular bisector of the segment connecting the two given points.

    Here's how we find it:

    1. Identify the given points and line.

       Let the two given points be $A(-5, 6)$ and $B(3, 2)$.

       Let the given line be $L_1: 4x - y - 2 = 0$.

       We are looking for a point $P(x, y)$ that lies on $L_1$ and satisfies $PA = PB$.

    2. Find the equation of the perpendicular bisector of the segment $AB$.

       The perpendicular bisector is a line that passes through the midpoint of $AB$ and is perpendicular to $AB$.

       • Calculate the midpoint $M$ of $AB$.

         The coordinates of the midpoint $M(x_m, y_m)$ of a segment with endpoints $(x_1, y_1)$ and $(x_2, y_2)$ are given by $x_m = \frac{x_1 + x_2}{2}$ and $y_m = \frac{y_1 + y_2}{2}$.

         For $A(-5, 6)$ and $B(3, 2)$:

         $x_m = \frac{-5 + 3}{2} = \frac{-2}{2} = -1$

         $y_m = \frac{6 + 2}{2} = \frac{8}{2} = 4$

         So, the midpoint is $M(-1, 4)$.

       • Calculate the slope of the segment $AB$.

         The slope $m_{AB}$ of a line passing through $(x_1, y_1)$ and $(x_2, y_2)$ is $m = \frac{y_2 - y_1}{x_2 - x_1}$.

         $m_{AB} = \frac{2 - 6}{3 - (-5)} = \frac{-4}{3 + 5} = \frac{-4}{8} = -\frac{1}{2}$

       • Calculate the slope of the perpendicular bisector ($L_2$).

         If two lines are perpendicular, the product of their slopes is $-1$.

         Let $m_{L_2}$ be the slope of the perpendicular bisector.

         $m_{L_2} \cdot m_{AB} = -1$

         $m_{L_2} \cdot \left(-\frac{1}{2}\right) = -1$

         $m_{L_2} = 2$

       • Form the equation of the perpendicular bisector ($L_2$). …

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