Q.Find the equation of lines passing through $(1,2)$ and making angle $30^\circ$ with $y$-axis.

Showing the 12 most recent of 61 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.If $\alpha, \beta \in \left( -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right), \cos^4 \alpha = \dfrac{1}{16}, \sin^4 \beta = \dfrac{1}{16}$ then $\cos \alpha + \cos \beta =$ (A) $\sqrt{2} \cos 15^\circ$ (B) $\sqrt{2} \sin 15^\circ$ (C) $-\sqrt{2} \cos 15^\circ$ (D) $-\sqrt{2} \sin 15^\circ$

   ›Reveal solutionSolution

   Given the fourth powers of cosine and sine, we extract the principal values of $\alpha$ and $\beta$ in $(-\pi/2,\pi/2)$, then compute $\cos\alpha+\cos\beta$ and match it to one of the given forms, obtaining $\sqrt{2}\cos15^\circ$.

   ---

   We start with the given equations:

   $\cos^4\alpha = \frac{1}{16}, \quad \sin^4\beta = \frac{1}{16}, \quad \alpha,\beta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right).$

   Concept and intuition:

   The fourth power equals $1/16$ means the absolute value of the base trig function is $1/2$ (since $(1/2)^4 = 1/16$). The interval $(-\pi/2,\pi/2)$ is where cosine is positive and sine is increasing from $-1$ to $1$, so we can uniquely determine the signs and angles. Then we just add the cosines and simplify to a known exact value.

   ---

   1. Extract $\cos\alpha$. From $\cos^4\alpha = 1/16$, taking the positive fourth root (since cosine is positive on $(-\pi/2,\pi/2)$) gives

   $|\cos\alpha| = \frac12 \quad\Rightarrow\quad \cos\alpha = \frac12.$

   The only angle in $(-\pi/2,\pi/2)$ with $\cos\alpha = 1/2$ is

   $\alpha = \frac{\pi}{3}.$

   2. Extract $\sin\beta$.

      From $\sin^4\beta = 1/16$, we get $|\sin\beta| = 1/2$. On $(-\pi/2,\pi/2)$, sine can be positive or negative. But note: $\sin\beta = \pm 1/2$ gives $\beta = \pm \pi/6$, both within the interval. However, we must check if any further condition restricts the sign — none is given, so both are possible. But the sum $\cos\alpha+\cos\beta$ will be the same for both because $\cos(\pi/6)=\cos(-\pi/6)=\sqrt{3}/2$. So we can take $\beta = \pi/6$ without loss.

   3. Compute the sum.

   $\cos\alpha + \cos\beta = \cos\frac{\pi}{3} + \cos\frac{\pi}{6} = \frac12 + \frac{\sqrt{3}}{2} = \frac{1+\sqrt{3}}{2}.$

   4. Match to the options. Options involve $\sqrt{2}\cos15^\circ$ or $\sqrt{2}\sin15^\circ$ (with possible minus signs). Recall: $$$$

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2. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.If $\overrightarrow{AB} = \hat{i} + \hat{j} - 2\hat{k}$, $\overrightarrow{CB} = 2\hat{i} - \hat{j} + a\hat{k}$ $(a \in \mathbb{Z})$ are two sides of a triangle ABC and the angle between these two sides is $\dfrac{\pi}{3}$, then the length of its third side is (A) $6$ (B) $2\sqrt{6}$ (C) $\sqrt{6}$ (D) $3\sqrt{6}$

   ›Reveal solutionSolution

   Use $\cos\dfrac{\pi}{3}=\dfrac12$ with the dot product to fix the integer $a=-1$, then the third side $\overrightarrow{AC}=\overrightarrow{AB}-\overrightarrow{CB}$ has length $\sqrt6$.

   Given $\overrightarrow{AB}=\hat i+\hat j-2\hat k$ and $\overrightarrow{CB}=2\hat i-\hat j+a\hat k$, with the angle between them $\dfrac{\pi}{3}$:

   $\overrightarrow{AB}\cdot\overrightarrow{CB}=2-1-2a=1-2a,\quad |\overrightarrow{AB}|=\sqrt6,\quad |\overrightarrow{CB}|=\sqrt{5+a^2}.$

   $\cos\frac{\pi}{3}=\frac{1-2a}{\sqrt6\,\sqrt{5+a^2}}=\frac12.$

   So $2(1-2a)=\sqrt6\,\sqrt{5+a^2}$ (requires $1-2a>0$). Squaring:

   $4(1-2a)^2=6(5+a^2)\ \Rightarrow\ 16a^2-16a+4=6a^2+30\ \Rightarrow\ 5a^2-8a-13=0.$ …

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3. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.If the direction cosines of the line common to the planes $x+2y-z-1=0$ and $3x-4y+z-5=0$ are $(l, m, n)$ then $|l+m-n|=$ (A) $\dfrac{6}{\sqrt{30}}$ (B) $\dfrac{4}{\sqrt{30}}$ (C) $\dfrac{2}{\sqrt{30}}$ (D) $\dfrac{8}{\sqrt{30}}$

   ›Reveal solutionSolution

   Direction of the common line is $\mathbf n_1\times\mathbf n_2\propto(1,2,5)$, so DCs $=\tfrac{1}{\sqrt{30}}(1,2,5)$ and $|l+m-n|=\tfrac{|1+2-5|}{\sqrt{30}}=\tfrac{2}{\sqrt{30}}$ — option (C).

   Direction. The common line lies in both planes, so its direction is the cross product of the normals $\mathbf n_1=(1,2,-1)$ and $\mathbf n_2=(3,-4,1):$

   $\mathbf n_1\times\mathbf n_2=\big(2(1)-(-1)(-4),\;-(1(1)-(-1)(3)),\;1(-4)-2(3)\big)=(-2,-4,-10)\propto(1,2,5).$ …

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4. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.Number of values of $\theta$ lying in the interval $(-\pi, \pi)$ such that

   [!FORMULA] $\left| \begin{matrix} \cos\theta & -\sin\theta & 1 \\ \sin\theta & 1 & -\cos\theta \\ 1 & \cos\theta & \sin\theta \end{matrix} \right| = 2$

   is (A) 1 (B) 0 (C) 2 (D) 3

   ›Reveal solutionSolution

   The determinant equals $\sin^3\theta + \cos^3\theta + 3\sin\theta\cos\theta - 1$, whose maximum value over all $\theta$ is only about $1.21 < 2$. So the equation $=2$ has no solution in $(-\pi,\pi)$. Option (B).

   Evaluating the determinant

   Expanding along the first row:

   $\Delta = \cos\theta\,(\sin\theta + \cos^2\theta) + \sin\theta\,(\sin^2\theta + \cos\theta) + (\sin\theta\cos\theta - 1).$

   Multiplying out:

   $\Delta = \sin\theta\cos\theta + \cos^3\theta + \sin^3\theta + \sin\theta\cos\theta + \sin\theta\cos\theta - 1,$

   $\Delta = \sin^3\theta + \cos^3\theta + 3\sin\theta\cos\theta - 1.$

   Solving $\Delta = 2$

   Let $s = \sin\theta + \cos\theta$ and $p = \sin\theta\cos\theta$, with $s^2 = 1 + 2p$, so $p = \dfrac{s^2-1}{2}$, and $\sin^3\theta+\cos^3\theta = s(1-p)$. Then

   $\Delta = s(1-p) + 3p - 1 = s + p(3-s) - 1.$

   Setting $\Delta = 2$ and substituting $p = \dfrac{s^2-1}{2}$ gives

   $2s + (s^2-1)(3-s) = 6 \;\Longrightarrow\; s^3 - 3s^2 - 3s + 9 = 0.$

   Factor by grouping:

   $s^2(s-3) - 3(s-3) = (s-3)(s^2-3) = 0 \;\Longrightarrow\; s = 3 \text{ or } s = \pm\sqrt3.$

   Feasibility of $s$ …

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5. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.If the length of the chord $x + y - 1 = 0$ of the circle $x^2 + y^2 - 6x + 2fy - 2 = 0$ $(f > 0)$ is $\dfrac{6\sqrt{7}}{\sqrt{2}}$, then the length of the intercept made by this circle on Y-axis is (A) $3\sqrt{3}$ (B) $12\sqrt{3}$ (C) $6\sqrt{3}$ (D) $4\sqrt{3}$

   ›Reveal solutionSolution

   The key idea is to use the chord-length formula for a line intersecting a circle, which relates the perpendicular distance from the centre to the chord and the radius. Solving for $f$ gives the Y-intercept length as $6\sqrt{3}$.

   The problem gives a chord of a circle with a known length, and asks for the Y-intercept of the same circle. The Y-intercept is simply the length of the chord the circle cuts on the Y-axis — that is, the distance between the two points where $x=0$ meets the circle. So we need the circle’s radius and centre first, which come from the given chord condition.

   Concept: For a circle of radius $r$ and a chord at perpendicular distance $d$ from the centre, the chord length is $2\sqrt{r^2 - d^2}$. This is a direct consequence of the right triangle formed by the radius to an endpoint, the perpendicular from centre to chord, and half the chord.

   We are told the chord $x + y - 1 = 0$ has length $\frac{6\sqrt{7}}{\sqrt{2}}$. We’ll find $f$ by equating this to $2\sqrt{r^2 - d^2}$.

   1. Find the centre and radius of the circle.

      The circle is $x^2 + y^2 - 6x + 2fy - 2 = 0$.

      Complete squares:

      $x^2 - 6x = (x-3)^2 - 9$

      $y^2 + 2fy = (y+f)^2 - f^2$

      So the equation becomes

      $(x-3)^2 + (y+f)^2 = 9 + f^2 + 2 = f^2 + 11$

      Hence centre $C = (3, -f)$ and radius $r = \sqrt{f^2 + 11}$.

   2. Perpendicular distance from centre to the chord.

      The chord line is $x + y - 1 = 0$.

      Distance from $C(3, -f)$ to this line:

   $d = \frac{|3 + (-f) - 1|}{\sqrt{1^2 + 1^2}} = \frac{|2 - f|}{\sqrt{2}}$

   Since $f > 0$, we keep the absolute value — it will be resolved by squaring later.

   3. Apply the chord-length formula. Chord length $= 2\sqrt{r^2 - d^2} = \frac{6\sqrt{7}}{\sqrt{2}}$. Square both sides:

   $4(r^2 - d^2) = \frac{36 \cdot 7}{2} = 126$

   So $r^2 - d^2 = \frac{126}{4} = \frac{63}{2}$.

   Substitute $r^2 = f^2 + 11$ and $d^2 = \frac{(2-f)^2}{2}$:

   $f^2 + 11 - \frac{(2-f)^2}{2} = \frac{63}{2}$

   Multiply through by 2:

   $2f^2 + 22 - (4 - 4f + f^2) = 63$

   Simplify: $2f^2 + 22 - 4 + 4f - f^2 = 63$

   $\Rightarrow f^2 + 4f + 18 = 63$

   $\Rightarrow f^2 + 4f - 45 = 0$ …

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6. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.Let a tangent $L_1$ with slope $m$ drawn to the parabola $y^2 = 8x$ be perpendicular to a normal $L_2$ drawn to the parabola $y^2 = 12x$. If $m = 1$ and the point of intersection of $L_1$ and $L_2$ is $(\alpha, \beta)$ then $\alpha + \beta =$ (A) $9$ (B) $3$ (C) $6$ (D) $12$

   ›Reveal solutionSolution

   We find the equation of the tangent $L_1$ to $y^2=8x$ with slope $m=1$, then determine the slope of the normal $L_2$ to $y^2=12x$ using the perpendicularity condition. Finally, we find the intersection of $L_1$ and $L_2$ and sum its coordinates to get $\boxed{9}$.

   To solve this problem, we need to recall the standard forms for the equations of a tangent and a normal to a parabola $y^2 = 4ax$. The problem involves two different parabolas, so we must be careful to use the correct parameter $a$ for each.

   The general equation of a tangent to the parabola $y^2 = 4ax$ with slope $m$ is given by:

   $y = mx + \frac{a}{m}$

   The general equation of a normal to the parabola $y^2 = 4ax$ with slope $m'$ is given by:

   $y = m'x - 2am' - am'^3$

   We are given the slope of the tangent $L_1$ and the condition that $L_1$ is perpendicular to the normal $L_2$. This perpendicularity condition will allow us to find the slope of $L_2$. Once we have the equations for both lines, we can find their point of intersection and sum its coordinates.

   Here is the step-by-step solution:

   1. Determine the equation of the tangent $L_1$:

      The first parabola is $y^2 = 8x$. Comparing this with the standard form $y^2 = 4ax$, we find $4a = 8$, which means $a_1 = 2$.

      The slope of the tangent $L_1$ is given as $m = 1$.

      Using the formula for the tangent $y = mx + \frac{a}{m}$, we substitute $m=1$ and $a_1=2$:

      $L_1: y = (1)x + \frac{2}{1}$

      $L_1: y = x + 2$

   2. Determine the slope of the normal $L_2$:

      We are given that the tangent $L_1$ is perpendicular to the normal $L_2$.

      The slope of $L_1$ is $m_1 = 1$.

      If two lines are perpendicular, the product of their slopes is $-1$. Let the slope of $L_2$ be $m_2$.

      $m_1 m_2 = -1$

      $(1) m_2 = -1$

      $m_2 = -1$

      So, the slope of the normal $L_2$ is $-1$.

   3. Determine the equation of the normal $L_2$:

      The second parabola is $y^2 = 12x$. Comparing this with $y^2 = 4ax$, we find $4a = 12$, which means $a_2 = 3$.

      The slope of the normal $L_2$ is $m_2 = -1$. …

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7. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.Let $S \equiv x^2 + y^2 - 6x + 4y + c = 0$, $S' \equiv x^2 + y^2 - 4x + 6y + 9 = 0$, $S'' \equiv x^2 + y^2 + 5x + 3y + k = 0$ be three circles. If the angles of intersection of the circle $S' = 0$ with the circles $S = 0$ and $S'' = 0$ are respectively $\dfrac{\pi}{4}$ and $\dfrac{\pi}{2}$, then $c + k =$ (A) $1$ (B) $15$ (C) $21$ (D) $9$

   ›Reveal solutionSolution

   The key idea is to use the formula for the angle between two circles: $\cos\theta = \frac{d^2 - r_1^2 - r_2^2}{2 r_1 r_2}$.

   Applying it to the given angles yields two equations in $c$ and $k$, solving which gives $c + k = 15$.

   We have three circles given by their equations:

   $\begin{aligned} S &: x^2 + y^2 - 6x + 4y + c = 0, \\ S' &: x^2 + y^2 - 4x + 6y + 9 = 0, \\ S'' &: x^2 + y^2 + 5x + 3y + k = 0. \end{aligned}$

   The angle of intersection between two circles is defined as the angle between their tangents at a point of intersection. It can be computed from the radii and the distance between centers using the cosine rule.

   ---

   1. Find centers and radii

   Rewrite each circle in center-radius form $(x - h)^2 + (y - k)^2 = r^2$.

   • For $S$:

     $x^2 - 6x + y^2 + 4y + c = 0$

     Complete squares: $(x-3)^2 - 9 + (y+2)^2 - 4 + c = 0$

     $\Rightarrow (x-3)^2 + (y+2)^2 = 13 - c$

     So center $C_1 = (3, -2)$, radius $r_1 = \sqrt{13 - c}$ (requires $c < 13$ for a real circle).

   • For $S'$:

     $x^2 - 4x + y^2 + 6y + 9 = 0$

     Complete squares: $(x-2)^2 - 4 + (y+3)^2 - 9 + 9 = 0$

     $\Rightarrow (x-2)^2 + (y+3)^2 = 4$

     So center $C_2 = (2, -3)$, radius $r_2 = 2$.

   • For $S''$:

     $x^2 + 5x + y^2 + 3y + k = 0$

     Complete squares: $(x + \frac{5}{2})^2 - \frac{25}{4} + (y + \frac{3}{2})^2 - \frac{9}{4} + k = 0$

     $\Rightarrow (x + \frac{5}{2})^2 + (y + \frac{3}{2})^2 = \frac{34}{4} - k = \frac{17}{2} - k$

     So center $C_3 = \left(-\frac{5}{2}, -\frac{3}{2}\right)$, radius $r_3 = \sqrt{\frac{17}{2} - k}$.

   ---

   2. Distance between centers

   • Distance $d_{12}$ between $C_1(3,-2)$ and $C_2(2,-3)$:

     $d_{12} = \sqrt{(3-2)^2 + (-2+3)^2} = \sqrt{1 + 1} = \sqrt{2}$.

   • Distance $d_{23}$ between $C_2(2,-3)$ and $C_3\left(-\frac{5}{2}, -\frac{3}{2}\right)$:

     $d_{23} = \sqrt{\left(2 + \frac{5}{2}\right)^2 + \left(-3 + \frac{3}{2}\right)^2} = \sqrt{\left(\frac{9}{2}\right)^2 + \left(-\frac{3}{2}\right)^2} = \sqrt{\frac{81}{4} + \frac{9}{4}} = \sqrt{\frac{90}{4}} = \frac{\sqrt{90}}{2} = \frac{3\sqrt{10}}{2}$.

   ---

   3. Apply angle formula

   The angle $\theta$ between two circles with radii $r_a$, $r_b$ and center distance $d$ satisfies:

   $\cos\theta = \frac{d^2 - r_a^2 - r_b^2}{2 r_a r_b}.$

   First condition: $\theta = \frac{\pi}{4}$ between $S$ and $S'$.

   $\cos\frac{\pi}{4} = \frac{\sqrt{2}}{2} = \frac{d_{12}^2 - r_1^2 - r_2^2}{2 r_1 r_2}.$

   Substitute $d_{12}^2 = 2$, $r_2 = 2$, $r_1^2 = 13 - c$:

   $\frac{\sqrt{2}}{2} = \frac{2 - (13 - c) - 4}{2 \cdot \sqrt{13 - c} \cdot 2}.$ …

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8. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.The radical axis of the circles $x^2 + y^2 + 4x + 6y + 7 = 0$ and $4x^2 + 4y^2 + 8x + 12y - 24 = 0$ is a tangent to the circle $x^2 + y^2 = 13$ at a point $(\alpha, \beta)$, then $\alpha + \beta =$ (A) $0$ (B) $-\dfrac{5}{2}$ (C) $1$ (D) $-5$

   ›Reveal solutionSolution

   The radical axis is the line obtained by subtracting the equations of the two circles. After simplifying, we find it is $x + y + 5 = 0$. This line is tangent to $x^2 + y^2 = 13$ at $(\alpha, \beta)$, so the radius to the point of tangency is perpendicular to the line, giving $\alpha = \beta$ and then $\alpha^2 + \beta^2 = 13$ yields $\alpha = \beta = \pm \sqrt{13/2}$. Checking which point lies on the line gives $(\alpha, \beta) = \left(-\frac{5}{2}, -\frac{5}{2}\right)$, so $\alpha + \beta = -5$. The correct option is (D).

   ---

   Concept & Intuition

   The radical axis of two circles is the set of points having equal power with respect to both circles. For circles written in standard form, it’s simply the line you get by subtracting one equation from the other (after making the coefficients of $x^2$ and $y^2$ the same). Once we have that line, the condition “it is tangent to a given circle” means the distance from the circle’s center to the line equals the radius. Moreover, the point of tangency lies on both the line and the circle, and the radius to that point is perpendicular to the tangent line. That perpendicularity gives a simple relation between $\alpha$ and $\beta$.

   ---

   Step-by-step solution

   1. Write both circles with matching quadratic coefficients. The first circle is $x^2 + y^2 + 4x + 6y + 7 = 0$. The second is $4x^2 + 4y^2 + 8x + 12y - 24 = 0$. Divide the second equation by $4$ to get

   $x^2 + y^2 + 2x + 3y - 6 = 0.$

   2. Find the radical axis. Subtract the second (adjusted) equation from the first:

   $(x^2 + y^2 + 4x + 6y + 7) - (x^2 + y^2 + 2x + 3y - 6) = 0 - 0$

   Simplifying:

   $(4x - 2x) + (6y - 3y) + (7 + 6) = 0 \quad\Rightarrow\quad 2x + 3y + 13 = 0.$

   So the radical axis is the line

   $2x + 3y + 13 = 0.$

   3. Check tangency condition to $x^2 + y^2 = 13$. The circle $x^2 + y^2 = 13$ has center $(0,0)$ and radius $r = \sqrt{13}$. For the line $2x + 3y + 13 = 0$ to be tangent, the distance from $(0,0)$ to the line must equal $\sqrt{13}$. Distance formula:

   $\frac{|2\cdot0 + 3\cdot0 + 13|}{\sqrt{2^2 + 3^2}} = \frac{13}{\sqrt{13}} = \sqrt{13}.$

   It matches exactly — so the line is indeed tangent.

   4. Find the point of tangency $(\alpha, \beta)$. The radius $OP$ (from center $(0,0)$ to $(\alpha, \beta)$) is perpendicular to the tangent line. The line’s normal vector is $(2,3)$, so the radius vector $(\alpha, \beta)$ must be parallel to $(2,3)$. Hence …

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9. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.Let P, Q, R, S be the points of intersection of the circle $x^2 + y^2 = 4$ and the hyperbola $xy = \sqrt{3}$. If $P = (\alpha, \beta)$ and $\alpha > \beta > 0$, then the equation of the tangent drawn at P to the hyperbola is (A) $x + y = 2$ (B) $x + \sqrt{3}y = 2\sqrt{3}$ (C) $\sqrt{3}x + y = \sqrt{3}$ (D) $x - y = 0$

   ›Reveal solutionSolution

   The key idea is to find the intersection point $P$ in the first quadrant where the circle and hyperbola meet, then use the derivative of the hyperbola to write its tangent line equation. The correct option is (B).

   We are given two curves:

   • Circle: $x^2 + y^2 = 4$
   • Hyperbola: $xy = \sqrt{3}$

   They intersect at four points. We are told $P = (\alpha, \beta)$ with $\alpha > \beta > 0$, so $P$ is in the first quadrant and closer to the x-axis than the y-axis.

   ---

   1. Find the intersection point $P$ in the first quadrant

   We solve the system:

   $x^2 + y^2 = 4, \quad xy = \sqrt{3}.$

   A classic trick: square the second equation and add to the first in a useful way.

   From $xy = \sqrt{3}$, we have $x^2 y^2 = 3$.

   Now consider $(x + y)^2 = x^2 + y^2 + 2xy = 4 + 2\sqrt{3}$.

   And $(x - y)^2 = x^2 + y^2 - 2xy = 4 - 2\sqrt{3}$.

   Since $\alpha > \beta > 0$, we have $x - y > 0$, so:

   $x - y = \sqrt{4 - 2\sqrt{3}}.$

   Simplify: $4 - 2\sqrt{3} = (\sqrt{3} - 1)^2$ because $(\sqrt{3} - 1)^2 = 3 + 1 - 2\sqrt{3} = 4 - 2\sqrt{3}$.

   Thus:

   $x - y = \sqrt{3} - 1.$

   Also:

   $x + y = \sqrt{4 + 2\sqrt{3}} = \sqrt{(\sqrt{3} + 1)^2} = \sqrt{3} + 1.$

   Now solve:

   $x = \frac{(x+y)+(x-y)}{2} = \frac{(\sqrt{3}+1)+(\sqrt{3}-1)}{2} = \frac{2\sqrt{3}}{2} = \sqrt{3},$

   $y = \frac{(x+y)-(x-y)}{2} = \frac{(\sqrt{3}+1)-(\sqrt{3}-1)}{2} = \frac{2}{2} = 1.$

   So $P = (\sqrt{3}, 1)$. Indeed $\alpha = \sqrt{3} > \beta = 1 > 0$, matching the condition.

   ---

   2. Find the tangent to the hyperbola at $P$

   The hyperbola is $xy = \sqrt{3}$. Differentiate implicitly:

   $y + x \frac{dy}{dx} = 0 \quad \Rightarrow \quad \frac{dy}{dx} = -\frac{y}{x}.$ …

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10. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
    Q.Let P, Q, R, S be the points of intersection of the circle $x^2 + y^2 = 4$ and the hyperbola $xy = \sqrt{3}$. If $P = (\alpha, \beta)$ and $\alpha > \beta > 0$, then the equation of the tangent drawn at P to the hyperbola is (A) $x - y = 0$ (B) $\sqrt{3}x + y = \sqrt{3}$ (C) $x + \sqrt{3}y = 2\sqrt{3}$ (D) $x + y = 2$

    ›Reveal solutionSolution

    The key is to find the intersection point $P$ in the first quadrant where both curves meet, then use the derivative of the hyperbola to write its tangent line. The correct tangent equation is $x + \sqrt{3}y = 2\sqrt{3}$, which corresponds to option (C).

    We have the circle $x^2 + y^2 = 4$ and the rectangular hyperbola $xy = \sqrt{3}$. Their intersection points are symmetric; we are told $P = (\alpha, \beta)$ with $\alpha > \beta > 0$, so $P$ lies in the first quadrant and closer to the x-axis than the y-axis.

    ---

    1. Find the coordinates of $P$

    We solve the system:

    $x^2 + y^2 = 4, \quad xy = \sqrt{3}.$

    A classic trick: square the second equation and add/subtract.

    From $xy = \sqrt{3}$, we have $x^2 y^2 = 3$.

    Now consider $(x^2 + y^2)^2 = x^4 + 2x^2 y^2 + y^4 = 16$.

    Substitute $x^2 y^2 = 3$:

    $x^4 + y^4 + 6 = 16 \quad\Rightarrow\quad x^4 + y^4 = 10.$

    Also, $(x^2 - y^2)^2 = x^4 + y^4 - 2x^2 y^2 = 10 - 6 = 4$.

    Thus $x^2 - y^2 = \pm 2$. Since $\alpha > \beta > 0$, we have $\alpha^2 > \beta^2$, so

    $\alpha^2 - \beta^2 = 2.$

    Now solve:

    $\alpha^2 + \beta^2 = 4,\quad \alpha^2 - \beta^2 = 2.$

    Adding: $2\alpha^2 = 6 \Rightarrow \alpha^2 = 3 \Rightarrow \alpha = \sqrt{3}$ (positive).

    Subtracting: $2\beta^2 = 2 \Rightarrow \beta^2 = 1 \Rightarrow \beta = 1$.

    So $P = (\sqrt{3}, 1)$.

    Tip

    Instead of squaring, you could also substitute $y = \sqrt{3}/x$ into the circle: $x^2 + 3/x^2 = 4$, multiply by $x^2$ to get $x^4 - 4x^2 + 3 = 0$, factor as $(x^2 - 1)(x^2 - 3) = 0$. Since $\alpha > \beta > 0$, we pick $x = \sqrt{3}$, then $y = 1$.

    ---

    2. Equation of the tangent to the hyperbola at $P$

    The hyperbola is $xy = \sqrt{3}$. Differentiate implicitly:

    $$$$

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11. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
    Q.A plane $\pi_1$ contains the vectors $\vec{i} + \vec{j}$ and $\vec{i} + 2\vec{j}$. Another plane $\pi_2$ contains the vectors $2\vec{i} - \vec{j}$ and $3\vec{i} + 2\vec{k}$. $\vec{a}$ is a vector parallel to the line of intersection of $\pi_1$ and $\pi_2$. If the angle $\theta$ between $\vec{a}$ and $\vec{i} - 2\vec{j} + 2\vec{k}$ is acute, then $\theta =$ (A) $\dfrac{\pi}{2}$ (B) $\dfrac{\pi}{4}$ (C) $\cos^{-1} \left( \dfrac{4}{3\sqrt{5}} \right)$ (D) $\cos^{-1} \left( \dfrac{2}{\sqrt{5}} \right)$

    ›Reveal solutionSolution

    The line of intersection of two planes is perpendicular to both normal vectors; we find the cross product of the normals, then compute the acute angle with the given vector, matching one of the options.

    Concept and intuition:

    The line of intersection of two planes is the set of points lying in both planes. A direction vector of this line must be perpendicular to the normal of each plane (since it lies in both planes). Therefore, the direction vector is parallel to the cross product of the two normals. Once we find that direction, we compute the angle between it and the given vector, ensuring the acute angle is chosen.

    ---

    1. Find a normal to plane $\pi_1$. $\pi_1$ contains $\vec{v}_1 = \vec{i} + \vec{j}$ and $\vec{v}_2 = \vec{i} + 2\vec{j}$. A normal $\vec{n}_1$ is their cross product:

    $\vec{n}_1 = \vec{v}_1 \times \vec{v}_2 = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 1 & 0 \\ 1 & 2 & 0 \end{vmatrix} = (1\cdot0 - 0\cdot2)\vec{i} - (1\cdot0 - 0\cdot1)\vec{j} + (1\cdot2 - 1\cdot1)\vec{k} = 0\vec{i} - 0\vec{j} + 1\vec{k} = \vec{k}.$

    So $\vec{n}_1 = \vec{k}$.

    2. Find a normal to plane $\pi_2$. $\pi_2$ contains $\vec{w}_1 = 2\vec{i} - \vec{j}$ and $\vec{w}_2 = 3\vec{i} + 2\vec{k}$. Their cross product:

    $\vec{n}_2 = \vec{w}_1 \times \vec{w}_2 = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & -1 & 0 \\ 3 & 0 & 2 \end{vmatrix} = ((-1)\cdot2 - 0\cdot0)\vec{i} - (2\cdot2 - 0\cdot3)\vec{j} + (2\cdot0 - (-1)\cdot3)\vec{k}$

    $= (-2)\vec{i} - (4)\vec{j} + (3)\vec{k} = -2\vec{i} - 4\vec{j} + 3\vec{k}.$

    3. Direction of the line of intersection. A vector $\vec{a}$ parallel to the intersection is perpendicular to both normals, so

    $\vec{a} \parallel \vec{n}_1 \times \vec{n}_2.$

    Compute:

    $\vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 0 & 0 & 1 \\ -2 & -4 & 3 \end{vmatrix} = (0\cdot3 - 1\cdot(-4))\vec{i} - (0\cdot3 - 1\cdot(-2))\vec{j} + (0\cdot(-4) - 0\cdot(-2))\vec{k}$

    $= 4\vec{i} - (2)\vec{j} + 0\vec{k} = 4\vec{i} - 2\vec{j}.$

    So we can take $\vec{a} = 4\vec{i} - 2\vec{j}$ (or any scalar multiple). …

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12. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
    Q.Two non parallel sides of a rhombus are parallel to the lines $x + y - 1 = 0$ and $7x - y - 5 = 0$. If $(1,3)$ is the centre of the rhombus and one of its vertices $A(\alpha, \beta)$ lies on $15x - 5y = 6$, then one of the possible values of $(\alpha + \beta)$ is (A) $\dfrac{18}{5}$ (B) $\dfrac{12}{5}$ (C) $\dfrac{37}{5}$ (D) $\dfrac{39}{5}$

    ›Reveal solutionSolution

    The rhombus’s sides are parallel to two given lines; its centre is known, and one vertex lies on a given line. Using the fact that the diagonals of a rhombus are perpendicular and bisect each other, we find the possible vertices and compute $\alpha+\beta$. The correct option is (D).

    We are told that two non‑parallel sides of a rhombus are parallel to the lines

    $x+y-1=0 \quad\text{and}\quad 7x-y-5=0.$

    Thus the sides of the rhombus have slopes $-1$ and $7$. The centre (intersection of the diagonals) is $(1,3)$. One vertex $A(\alpha,\beta)$ lies on the line $15x-5y=6$, i.e. $3x-y=\frac{6}{5}$. We need a possible value of $\alpha+\beta$.

    Key idea: In a rhombus, the diagonals are perpendicular and bisect each other. The sides are parallel to the given lines, so the diagonals are along the angle bisectors of those directions. Using the centre, we can find the equations of the diagonals, then intersect them with the side‑direction lines to locate vertices.

    ---

    1. Find the slopes of the diagonals. The sides have slopes $m_1=-1$ and $m_2=7$. The diagonals of a rhombus are the angle bisectors of the sides. The slopes of the bisectors satisfy

    $\frac{m - m_1}{1 + m m_1} = \pm \frac{m - m_2}{1 + m m_2}.$

    For $m_1=-1$, $m_2=7$:

    $\frac{m+1}{1 - m} = \pm \frac{m-7}{1+7m}.$

    Taking the $+$ sign:

    $(m+1)(1+7m) = (m-7)(1-m).$

    Expanding:

    $m+7m^2+1+7m = m - m^2 -7 +7m$

    $\Rightarrow 7m^2+8m+1 = -m^2+8m-7$

    $\Rightarrow 8m^2 = -8$ → no real solution.

    Taking the $-$ sign:

    $(m+1)(1+7m) = -(m-7)(1-m).$

    Left: $7m^2+8m+1$. Right: $-(m-7)(1-m) = (m-7)(m-1) = m^2 -8m +7$.

    So

    $7m^2+8m+1 = m^2 -8m +7 \quad\Rightarrow\quad 6m^2+16m-6=0.$

    Divide by 2: $3m^2+8m-3=0$.

    Solve: $m = \frac{-8 \pm \sqrt{64+36}}{6} = \frac{-8 \pm 10}{6}$.

    Hence $m = \frac{2}{6} = \frac13$ or $m = \frac{-18}{6} = -3$.

    So the diagonals have slopes $\frac13$ and $-3$ (they are perpendicular, as expected).

    2. Equations of the diagonals through the centre $(1,3)$. Diagonal 1 (slope $\frac13$):

    $y-3 = \frac13 (x-1) \quad\Rightarrow\quad x - 3y + 8 = 0.$

    Diagonal 2 (slope $-3$):

    $y-3 = -3(x-1) \quad\Rightarrow\quad 3x + y - 6 = 0.$

    3. Find the vertices.

       The sides are parallel to the given lines. So through each vertex, two sides run: one parallel to $x+y-1=0$ (slope $-1$) and one parallel to $7x-y-5=0$ (slope $7$). The diagonals intersect at the centre, and each diagonal connects opposite vertices.

       Let’s find the intersection of diagonal 1 with lines through the centre that are parallel to the sides — but more directly:

       A vertex lies on a diagonal and also on a line through the centre parallel to a side? No — the centre is the midpoint of the diagonal, not necessarily on a side. Better: The vertices are the intersections of lines through the centre that are parallel to the sides? Actually, the sides themselves are offset from the centre.

       Classic method: The diagonals are the angle bisectors. The sides are lines parallel to the given lines. The centre is the intersection of the diagonals. The vertices are the points where a line through the centre parallel to one diagonal meets lines parallel to the sides? That’s messy.

       Instead, use vector approach: Let the direction vectors of the sides be

    $\mathbf{u} = (1,-1) \quad\text{(slope $-1$)}, \qquad \mathbf{v} = (1,7) \quad\text{(slope $7$)}.$

    From the centre $O=(1,3)$, the vertices are at

    $O \pm a\mathbf{u} \pm b\mathbf{v}$

    for some scalars $a,b$. But the diagonals are along $\mathbf{u}+\mathbf{v}$ and $\mathbf{u}-\mathbf{v}$ (since diagonals are sums/differences of side vectors). Indeed,

    $\mathbf{u}+\mathbf{v} = (2,6) \parallel (1,3) \text{ (slope 3? Wait, slope 3, not 1/3)}.$

    Check: $\mathbf{u}+\mathbf{v}=(2,6)$ has slope $3$, but we need slope $1/3$ or $-3$. So maybe we need to scale: Actually, the diagonals are along the angle bisectors, which are $\frac{\mathbf{u}}{|\mathbf{u}|} \pm \frac{\mathbf{v}}{|\mathbf{v}|}$.

    $|\mathbf{u}| = \sqrt{2}$, $|\mathbf{v}| = \sqrt{50}=5\sqrt{2}$.

    So

    $\frac{\mathbf{u}}{\sqrt{2}} \pm \frac{\mathbf{v}}{5\sqrt{2}} = \frac{1}{5\sqrt{2}}(5\mathbf{u} \pm \mathbf{v}).$

    $5\mathbf{u} = (5,-5)$, so

    $5\mathbf{u} + \mathbf{v} = (6,2) \parallel (3,1) \text{ (slope $1/3$)},$

    $5\mathbf{u} - \mathbf{v} = (4,-12) \parallel (1,-3) \text{ (slope $-3$)}.$ …

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