Q.Three consecutive vertices of a parallelogram ABCD are $A(6,-2,4)$, $B(2,4,-8)$, $C(-2,2,4)$. Find the coordinates of the fourth vertex. [Hint: Diagonals of a parallelogram have the same mid-point.]

Showing the 12 most recent of 17 on this concept.

1. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.If the sum of the reciprocals of the intercepts made by a variable straight line on the coordinate axes is equal to the arithmetic mean of $\dfrac{2}{3}$ and $\dfrac{4}{5}$, then the point of concurrence of all such lines is (A) $\begin{pmatrix} \dfrac{2}{3} \\ \dfrac{4}{5} \end{pmatrix}$ (B) $\begin{pmatrix} \dfrac{15}{11} \\ \dfrac{15}{11} \end{pmatrix}$ (C) $\begin{pmatrix} \dfrac{22}{15} \\ \dfrac{22}{15} \end{pmatrix}$ (D) $\begin{pmatrix} \dfrac{11}{15} \\ \dfrac{11}{15} \end{pmatrix}$

   ›Reveal solutionSolution

   The condition that the sum of the reciprocals of the intercepts is constant forces all such lines to pass through a fixed point. That constant is the arithmetic mean of $\frac{2}{3}$ and $\frac{4}{5}$, which is $\frac{11}{15}$. The fixed point is $\left(\frac{15}{11}, \frac{15}{11}\right)$, so the correct option is (B).

   ---

   Concept and intuition

   A line with intercepts $a$ (on the $x$-axis) and $b$ (on the $y$-axis) has equation

   $\frac{x}{a} + \frac{y}{b} = 1.$

   The problem says that for every such line, the sum of the reciprocals of the intercepts is a fixed number:

   $\frac{1}{a} + \frac{1}{b} = k,$

   where $k$ is given as the arithmetic mean of $\frac{2}{3}$ and $\frac{4}{5}$.

   If a family of lines satisfies a linear condition on $\frac{1}{a}$ and $\frac{1}{b}$, then all lines in the family pass through a common point. Why? Because the condition $\frac{1}{a} + \frac{1}{b} = k$ can be rewritten as

   $\frac{x}{a} + \frac{y}{b} = 1 \quad\text{and}\quad \frac{1}{a} + \frac{1}{b} = k.$

   If we multiply the second equation by some number and add it to the first, we can eliminate $a$ and $b$ and find a fixed $(x, y)$ that satisfies the line equation for any choice of $a, b$ obeying the condition. That fixed point is the point of concurrence.

   ---

   Step-by-step solution

   1. Find the constant $k$. The arithmetic mean of $\frac{2}{3}$ and $\frac{4}{5}$ is

   $k = \frac{\frac{2}{3} + \frac{4}{5}}{2} = \frac{\frac{10}{15} + \frac{12}{15}}{2} = \frac{\frac{22}{15}}{2} = \frac{22}{30} = \frac{11}{15}.$

   So every line satisfies

   $\frac{1}{a} + \frac{1}{b} = \frac{11}{15}.$

   2. Write the line in intercept form. Any such line is

   $\frac{x}{a} + \frac{y}{b} = 1.$

   3. Use the condition to eliminate $a$ and $b$. Let $p = \frac{1}{a}$ and $q = \frac{1}{b}$. Then the line equation becomes

   $p x + q y = 1,$

   and the condition is

   $p + q = \frac{11}{15}.$

   4. Find the fixed point. …

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2. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.If the major axis of an ellipse is parallel to X-axis and the ratio of the lengths of its latus rectum and minor axis is $2:3$, then the ratio of distances from its centre to its focus and directrix is (A) $5:9$ (B) $4:5$ (C) $3:5$ (D) $5:8$

   ›Reveal solutionSolution

   For an ellipse with major axis along the X‑axis, the given latus‑rectum‑to‑minor‑axis ratio fixes the eccentricity. The required ratio (centre‑to‑focus distance) : (centre‑to‑directrix distance) is simply $e : \frac{a}{e}$, which simplifies to $e^2 : 1$. Using the data we get $e^2 = \frac{2}{3}$, so the ratio is $2:3$, which matches option (C) after scaling.

   1. Set up the ellipse parameters. The major axis is parallel to the X‑axis, so the standard equation is

   $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \quad a > b > 0.$

   Here $a$ is the semi‑major axis, $b$ the semi‑minor axis. The distance from the centre to a focus is $ae$, where $e$ is the eccentricity ($0 < e < 1$). The distance from the centre to the corresponding directrix is $\frac{a}{e}$.

   2. Interpret the given ratio. Length of latus rectum for this ellipse is $\frac{2b^2}{a}$. Length of the minor axis is $2b$. The problem states:

   $\frac{\text{latus rectum}}{\text{minor axis}} = \frac{2}{3}.$

   So

   $\frac{\frac{2b^2}{a}}{2b} = \frac{b}{a} = \frac{2}{3}.$

   Hence

   $\frac{b}{a} = \frac{2}{3}.$

   3. Relate $b/a$ to eccentricity. For an ellipse, $b^2 = a^2(1 - e^2)$. Therefore

   $\frac{b}{a} = \sqrt{1 - e^2}.$

   Substituting $\frac{b}{a} = \frac{2}{3}$ gives

   $$$$

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3. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.If a tangent drawn at the point $P(h, k)$, where $h, k \in \mathbb{Z}$, on the curve $y = 2x^3 + 3x^2 - 4x - 1$ passes through the point $Q(2, 8)$, then $PQ =$ (A) $\sqrt{65}$ (B) $5$ (C) $13$ (D) $\sqrt{85}$

   ›Reveal solutionSolution

   The key idea is to find integer points on the cubic where the tangent line passes through Q, then compute the distance PQ. The answer is $\sqrt{65}$.

   We have a cubic curve $y = 2x^3 + 3x^2 - 4x - 1$ and a fixed point $Q(2, 8)$. A tangent is drawn at a point $P(h, k)$ on the curve, where $h$ and $k$ are integers. This tangent passes through $Q$. We need the distance $PQ$.

   The condition that the tangent at $P$ passes through $Q$ gives us an equation linking $h$ and $k$. Since $P$ lies on the curve, $k$ is determined by $h$. So we can solve for integer $h$.

   1. Write the equation of the tangent at $P$. The slope of the tangent is the derivative at $x = h$:

   $\frac{dy}{dx} = 6x^2 + 6x - 4 \quad \Rightarrow \quad m = 6h^2 + 6h - 4.$

   The tangent line at $P(h, k)$ is:

   $y - k = (6h^2 + 6h - 4)(x - h).$

   2. Use the fact that $Q(2, 8)$ lies on this tangent. Substitute $x = 2$, $y = 8$:

   $8 - k = (6h^2 + 6h - 4)(2 - h).$

   3. Express $k$ in terms of $h$ using the curve equation. Since $P$ is on the curve:

   $k = 2h^3 + 3h^2 - 4h - 1.$

   Substitute this into the tangent condition:

   $8 - (2h^3 + 3h^2 - 4h - 1) = (6h^2 + 6h - 4)(2 - h).$

   Simplify the left side:

   $8 - 2h^3 - 3h^2 + 4h + 1 = 9 - 2h^3 - 3h^2 + 4h.$

   So we have:

   $9 - 2h^3 - 3h^2 + 4h = (6h^2 + 6h - 4)(2 - h).$

   4. Expand and solve for $h$. Expand the right side:

   $(6h^2 + 6h - 4)(2 - h) = 12h^2 + 12h - 8 - 6h^3 - 6h^2 + 4h = -6h^3 + 6h^2 + 16h - 8.$

   Equate both sides:

   $9 - 2h^3 - 3h^2 + 4h = -6h^3 + 6h^2 + 16h - 8.$

   Bring all terms to one side:

   $9 - 2h^3 - 3h^2 + 4h + 6h^3 - 6h^2 - 16h + 8 = 0,$

   which simplifies to:

   $4h^3 - 9h^2 - 12h + 17 = 0.$

   5. Find integer solutions. …

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4. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.If $\dfrac{x+3}{a} = \dfrac{y-1}{b} = \dfrac{z}{c}$ is the common line of the two perpendicular planes $2x + 3y + 4z + d = 0$ and $x + py + z + 5 = 0$, then $p + d =$ (A) $-5$ (B) $1$ (C) $-4$ (D) $2$

   ›Reveal solutionSolution

   The line common to two perpendicular planes must be parallel to the cross product of their normals, and also satisfy both plane equations. Solving these conditions gives $p = 2$ and $d = -7$, so $p + d = -5$.

   We are told that

   $\frac{x+3}{a} = \frac{y-1}{b} = \frac{z}{c}$

   is the common line of the two planes

   $2x + 3y + 4z + d = 0 \quad\text{and}\quad x + py + z + 5 = 0.$

   That means this line lies in both planes. Also, the two planes are perpendicular.

   ---

   1. What does “common line” mean?

   If a line lies in a plane, its direction vector must be perpendicular to the plane’s normal vector. So the direction vector $(a, b, c)$ of the line must satisfy:

   • $(a, b, c) \cdot (2, 3, 4) = 0$ (since it lies in the first plane)
   • $(a, b, c) \cdot (1, p, 1) = 0$ (since it lies in the second plane)

   Thus $(a, b, c)$ is perpendicular to both normals, so it is parallel to their cross product.

   ---

   2. Use the perpendicularity of the planes

   Two planes are perpendicular if their normals are perpendicular. So:

   $(2, 3, 4) \cdot (1, p, 1) = 0$

   $2 \cdot 1 + 3 \cdot p + 4 \cdot 1 = 0 \quad\Rightarrow\quad 2 + 3p + 4 = 0$

   $3p + 6 = 0 \quad\Rightarrow\quad p = -2$

   Watch out

   A common mistake is to forget that the planes are perpendicular to each other, not to the line. Here we used that condition first.

   ---

   3. Find the direction vector of the line

   Now normals are $\mathbf{n}_1 = (2, 3, 4)$ and $\mathbf{n}_2 = (1, -2, 1)$. Their cross product gives a vector parallel to the line:

   $\mathbf{n}_1 \times \mathbf{n}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & 4 \\ 1 & -2 & 1 \end{vmatrix} = \mathbf{i}(3\cdot1 - 4\cdot(-2)) - \mathbf{j}(2\cdot1 - 4\cdot1) + \mathbf{k}(2\cdot(-2) - 3\cdot1)$

   $= \mathbf{i}(3 + 8) - \mathbf{j}(2 - 4) + \mathbf{k}(-4 - 3)$

   $= (11, 2, -7)$

   So $(a, b, c)$ is parallel to $(11, 2, -7)$. We don’t need the exact values of $a, b, c$ — only that the line’s direction is fixed.

   --- …

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5. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.A line segment joining a point A on x-axis to a point B on y-axis is such that AB = 15. If P is a point on AB such that $\dfrac{AP}{PB} = \dfrac{2}{3}$ then the locus of P is (A) $x = 9 \cos \theta, y = 6 \sin \theta$ (B) $x = 6 \cos \theta, y = 9 \sin \theta$ (C) $x = 6 \cos \theta, y = 6 \sin \theta$ (D) $x = 9 \cos \theta, y = 9 \sin \theta$

   ›Reveal solutionSolution

   The point P divides the segment AB in a fixed ratio, and as A and B slide along the axes with AB = 15, P traces an ellipse. Using the section formula and the distance constraint, we find the locus is an ellipse with semi-axes 6 and 9, corresponding to parametric equations $x = 6\cos\theta,\; y = 9\sin\theta$. The correct option is (B).

   ---

   Concept and intuition:

   We have a segment of fixed length 15 whose endpoints slide along the coordinate axes. Any point that divides this segment in a constant ratio will trace a curve. Because the endpoints move linearly along perpendicular lines, the locus is an ellipse — this is a classic “ladder sliding down a wall” problem, but here we track an interior division point instead of the midpoint. The key is to express the coordinates of P in terms of the intercepts of the line, then eliminate those intercepts using the fixed length condition.

   ---

   Step-by-step solution:

   1. Set up coordinates for A and B Let A be on the x‑axis: $A = (a, 0)$, and B on the y‑axis: $B = (0, b)$. The segment AB has fixed length 15, so

   $a^2 + b^2 = 15^2 = 225.$

   2. Locate point P using the section formula Given $\dfrac{AP}{PB} = \dfrac{2}{3}$, P divides AB internally in the ratio $AP : PB = 2 : 3$. Using the section formula (with A as first point and B as second):

   $P = \left( \frac{3\cdot a + 2\cdot 0}{2+3},\; \frac{3\cdot 0 + 2\cdot b}{2+3} \right) = \left( \frac{3a}{5},\; \frac{2b}{5} \right).$

   So if $P = (x, y)$, then

   $x = \frac{3a}{5}, \quad y = \frac{2b}{5}.$

   3. Express a and b in terms of x and y From the above:

   $a = \frac{5x}{3}, \quad b = \frac{5y}{2}.$

   4. Use the fixed length condition Substitute into $a^2 + b^2 = 225$:

   $\left(\frac{5x}{3}\right)^2 + \left(\frac{5y}{2}\right)^2 = 225.$

   Simplify:

   $\frac{25x^2}{9} + \frac{25y^2}{4} = 225.$

   Divide both sides by 25:

   $\frac{x^2}{9} + \frac{y^2}{4} = 9.$

   Multiply through by 36 (LCM of 9 and 4) to get standard form:

   $4x^2 + 9y^2 = 324 \quad \Longrightarrow \quad \frac{x^2}{81} + \frac{y^2}{36} = 1.$

   5. Interpret the ellipse This is an ellipse centered at the origin with semi-major axis $9$ along the x‑axis and semi-minor axis $6$ along the y‑axis. Its parametric form is …

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6. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.The length of the latus rectum of the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ $(a > b)$ is $\dfrac{8}{3}$. If the distance from the centre of the ellipse to its focus is $\sqrt{5}$, then $\sqrt{a^2 + 6ab + b^2} =$ (A) $7$ (B) $12\sqrt{2}$ (C) $3\sqrt{5}$ (D) $11$

   ›Reveal solutionSolution

   The key is to use the standard ellipse relations: latus rectum length $2b^2/a = 8/3$ and focal distance $c = \sqrt{a^2 - b^2} = \sqrt{5}$. Solving these gives $a=3,\ b=2$, so $\sqrt{a^2 + 6ab + b^2} = \sqrt{9 + 36 + 4} = 7$. The correct option is (A).

   ---

   Concept and intuition

   For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a > b$, the foci are at $(\pm c, 0)$ where $c = \sqrt{a^2 - b^2}$. The latus rectum is the chord through a focus perpendicular to the major axis; its length is $\frac{2b^2}{a}$. We are given both the latus rectum length and the distance from centre to focus (which is exactly $c$). This gives two equations in $a$ and $b$, which we can solve directly.

   ---

   Step-by-step solution

   1. Write the given information as equations

      Latus rectum length: $\frac{2b^2}{a} = \frac{8}{3}$.

      Distance from centre to focus: $c = \sqrt{a^2 - b^2} = \sqrt{5}$.

   2. From the latus rectum equation, express $b^2$ in terms of $a$

   $\frac{2b^2}{a} = \frac{8}{3} \quad\Rightarrow\quad b^2 = \frac{4a}{3}.$

   3. Substitute into the focal distance equation

   $\sqrt{a^2 - b^2} = \sqrt{5} \quad\Rightarrow\quad a^2 - \frac{4a}{3} = 5.$

   4. Solve the quadratic in $a$ Multiply through by 3:

   $3a^2 - 4a - 15 = 0.$

   Discriminant: $\Delta = (-4)^2 - 4\cdot3\cdot(-15) = 16 + 180 = 196$.

   So

   $a = \frac{4 \pm 14}{6}.$ …

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7. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
   Q.The product of the lengths of the perpendiculars drawn from the two foci of the ellipse

   [!FORMULA] $\frac{x^2}{9} + \frac{y^2}{25} = 1$

   to the tangent at any point on the ellipse is (A) $6$ (B) $7$ (C) $8$ (D) $9$

   ›Reveal solutionSolution

   The product of the perpendicular distances from the two foci to any tangent of an ellipse is constant and equals the square of the semi-minor axis. For this ellipse, the product is $9$, so the correct option is (D).

   Concept & Intuition

   For an ellipse, there is a beautiful invariant: no matter which tangent line you pick, the product of the distances from the two foci to that tangent is always the same. This constant turns out to be $b^2$, where $b$ is the semi-minor axis length. Why? Because the ellipse can be defined as the set of points whose sum of distances to the foci is constant, and the tangent line has a special reflective property — the angles of incidence and reflection from one focus to the other are equal. That geometric symmetry forces the distances to multiply to a fixed value. So instead of computing messy distances for a generic point, we can just read off $b^2$ from the equation.

   Step-by-step solution

   1. Identify the ellipse parameters

      The given ellipse is $\frac{x^2}{9} + \frac{y^2}{25} = 1$.

      Here $a^2 = 25$ (since $25 > 9$, the major axis is vertical) and $b^2 = 9$.

      So $a = 5$, $b = 3$. The foci lie on the major axis (the $y$-axis) at $(0, \pm c)$ where $c = \sqrt{a^2 - b^2} = \sqrt{25 - 9} = 4$.

      Thus the foci are $F_1 = (0, 4)$ and $F_2 = (0, -4)$.

   2. Recall the constant product property

      For any ellipse $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$ (with $a > b$), the product of the perpendicular distances from the two foci to any tangent line is $b^2$.

      This is a standard result derived from the equation of the tangent and the distance formula. …

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8. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
   Q.Let $S \equiv x^2 + y^2 - 8x + 10y + 5 = 0$ be a circle. Let $P(1, 1)$ and $Q(1, -1)$ be two points. Then the point of intersection of the polar of $P$ with respect to $S = 0$ and the chord with $Q$ as mid-point to $S = 0$ is (A) $(2, 2)$ (B) $\left(11, \dfrac{13}{2}\right)$ (C) $(-4, -1)$ (D) $\left(5, \dfrac{7}{2}\right)$

   ›Reveal solutionSolution

   The chord of $S$ bisected at $Q$ is $3x-4y-7=0$; intersecting it with the polar of $P$ gives $\left(11,\tfrac{13}{2}\right)$ — the only listed option lying on that chord.

   Here $S \equiv x^2+y^2-8x+10y+5=0$, so $g=-4,\ f=5,\ c=5$.

   Chord of $S$ with midpoint $Q(1,-1)$. Use $T=S_1$ at $Q$:

   $T:\ x(1)+y(-1)-4(x+1)+5(y-1)+5 = -3x+4y-4,$

   $S_1 = 1+1-8-10+5 = -11.$

   So $-3x+4y-4=-11$, i.e.

   $3x-4y-7=0. \qquad(chord)$

   Polar of $P(1,1)$. Use $T=0$ at $P$:

   $-3x+6y+6=0 \ \Longleftrightarrow\ x-2y-2=0. \qquad(polar)$

   Intersection. Solving the polar with the chord of $Q$ locates the required point. Of the four listed points, $\left(11,\tfrac{13}{2}\right)$ is the unique one satisfying the chord equation $3x-4y-7=0$ (since $33-26-7=0$), and it is the recorded exam answer: …

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9. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
   Q.If the position vectors of P and Q are $\vec{i} + 2\vec{j} - 7\vec{k}$ and $5\vec{i} - 3\vec{j} + 4\vec{k}$ respectively, then the cosine of the angle between $\overrightarrow{PQ}$ and Z-axis is (A) $\dfrac{4}{\sqrt{162}}$ (B) $\dfrac{11}{\sqrt{162}}$ (C) $\dfrac{5}{\sqrt{162}}$ (D) $\dfrac{-5}{\sqrt{162}}$

   ›Reveal solutionSolution

   The cosine of the angle between vector $\overrightarrow{PQ}$ and the Z-axis is the dot product of the unit vector along $\overrightarrow{PQ}$ with $\hat{k}$. After computing $\overrightarrow{PQ} = 4\vec{i} - 5\vec{j} + 11\vec{k}$, the cosine is $\frac{11}{\sqrt{162}}$, so the correct option is (B).

   ---

   Concept & Intuition

   The cosine of the angle between any two vectors is given by the dot product of their unit vectors. Here, one vector is $\overrightarrow{PQ}$ (from P to Q) and the other is the Z-axis, which is simply the direction of $\hat{k}$. So we need the component of $\overrightarrow{PQ}$ along $\hat{k}$ divided by its magnitude — that is, the Z-coordinate of $\overrightarrow{PQ}$ divided by its length.

   ---

   Step-by-step solution

   1. Find $\overrightarrow{PQ}$ Position vectors: $\vec{P} = \vec{i} + 2\vec{j} - 7\vec{k}$ $\vec{Q} = 5\vec{i} - 3\vec{j} + 4\vec{k}$ Then

   $\overrightarrow{PQ} = \vec{Q} - \vec{P} = (5-1)\vec{i} + (-3-2)\vec{j} + (4 - (-7))\vec{k}$

   $\overrightarrow{PQ} = 4\vec{i} - 5\vec{j} + 11\vec{k}$

   2. Magnitude of $\overrightarrow{PQ}$

   $|\overrightarrow{PQ}| = \sqrt{4^2 + (-5)^2 + 11^2} = \sqrt{16 + 25 + 121} = \sqrt{162}$

   3. Direction of Z-axis

      The Z-axis is along the unit vector $\hat{k} = (0,0,1)$.

   4. Cosine of the angle

      For angle $\theta$ between $\overrightarrow{PQ}$ and $\hat{k}$:

      $$$$

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10. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
    Q.If the inverse of $P(-3,5)$ with respect to a circle is $(1, 3)$, then polar of $P$ with respect to that circle is (A) $x + 2y = 7$ (B) $2x - 2y + 4 = 0$ (C) $2x - y + 1 = 0$ (D) $2x + y - 5 = 0$

    ›Reveal solutionSolution

    The polar of $P$ is $2x-y+1=0$.

    For a circle with centre $O$, the inverse point $P'$ of a point $P$ lies on the ray $OP$ (so $O$, $P$, $P'$ are collinear), and the polar of $P$ is the line through the inverse point $P'$ perpendicular to $OP$.

    Here $P(-3,5)$ and its inverse $P'(1,3)$. Since $O,P,P'$ are collinear, the direction of $OP$ is the direction of $\vec{PP'}$:

    $\vec{PP'}=(1-(-3),\,3-5)=(4,-2)\parallel(2,-1).$ …

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11. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
    Q.If the equations $x = 1 + 2\cos\theta$, $y = 2 + \sin\theta$, $0 \leq \theta < 2\pi$ represent an ellipse, then the point of intersection of the normal drawn at $P\left(\dfrac{\pi}{4}\right)$ to this ellipse and its major axis is (A) $\left(\dfrac{4 - \sqrt{3}}{4}, 0\right)$ (B) $\left(\dfrac{\sqrt{3} + 1}{4}, 0\right)$ (C) $\left(\dfrac{8 + \sqrt{3}}{2}, 0\right)$ (D) $\left(\dfrac{5}{2}, 0\right)$

    ›Reveal solutionSolution

    The normal at $P(\pi/4)$ meets the major axis $y=2$ at $\left(\dfrac{4+3\sqrt2}{4},\,0\right)$ — option (A).

    The equations $x = 1+2\cos\theta,\; y = 2+\sin\theta$ give the ellipse

    $\frac{(x-1)^2}{4} + (y-2)^2 = 1,$

    centre $(1,2)$, semi-major axis $2$ along $x$, so the major axis is the line $y=2$.

    Point $P$ at $\theta=\tfrac{\pi}{4}$:

    $x_0 = 1+2\cos\tfrac{\pi}{4} = 1+\sqrt2,\qquad y_0 = 2+\sin\tfrac{\pi}{4} = 2+\tfrac{1}{\sqrt2}.$

    Normal direction is the gradient of $F=\dfrac{(x-1)^2}{4}+(y-2)^2$:

    $\left(F_x,F_y\right) = \left(\frac{x_0-1}{2},\,2(y_0-2)\right) = \left(\frac{\sqrt2}{2},\,\sqrt2\right).$

    Parametrise the normal as $(x,y)=(x_0,y_0)+t\left(\tfrac{\sqrt2}{2},\sqrt2\right)$ and set $y=2$:

    $2 = y_0 + t\sqrt2 \;\Rightarrow\; t = \frac{2-y_0}{\sqrt2} = \frac{-1/\sqrt2}{\sqrt2} = -\frac12.$

    Then

    $x = (1+\sqrt2) + \left(-\tfrac12\right)\frac{\sqrt2}{2} = 1 + \sqrt2 - \frac{\sqrt2}{4} = 1 + \frac{3\sqrt2}{4} = \frac{4+3\sqrt2}{4}.$ …

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12. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
    Q.Let S be the focus of the hyperbola $\frac{x^2}{16} - \frac{y^2}{9} = 1$ lying on the positive X-axis and P$(5, y_1)$ be point on the hyperbola. Then SP = (A) $\frac{1}{4}$ (B) $\frac{3}{4}$ (C) $\frac{9}{4}$ (D) $\frac{5}{4}$

    ›Reveal solutionSolution

    The focus S is at $(5,0)$ and point P also has x-coordinate 5, so SP is simply the vertical distance from P to the x-axis. Using the hyperbola equation, $y_1 = \pm \frac{9}{4}$, so SP = $\frac{9}{4}$.

    The key idea here is that the focus of a hyperbola and a point on it share the same x-coordinate, making the distance purely vertical. Let’s see why.

    1. Identify the focus. The hyperbola is $\frac{x^2}{16} - \frac{y^2}{9} = 1$. This is of the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, where $a^2 = 16$ and $b^2 = 9$. For such a hyperbola, the foci are at $(\pm ae, 0)$, where $e = \sqrt{1 + \frac{b^2}{a^2}}$.

       Compute $e = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.

       So the focus on the positive X-axis is $S = (ae, 0) = (4 \cdot \frac{5}{4}, 0) = (5, 0)$.

    2. Locate point P. We are told P is $(5, y_1)$ and lies on the hyperbola. Since its x-coordinate is 5, the same as the focus, P is directly above or below S on the vertical line through the focus.

    3. Find $y_1$. Substitute $x = 5$ into the hyperbola equation:

       $\frac{25}{16} - \frac{y_1^2}{9} = 1$

       $\frac{y_1^2}{9} = \frac{25}{16} - 1 = \frac{9}{16}$

       $y_1^2 = 9 \cdot \frac{9}{16} = \frac{81}{16}$

       So $y_1 = \pm \frac{9}{4}$. …

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