Q.The solution of the differential equation 𝑥𝑑𝑥 + 𝑦𝑑𝑦 = 0 represents a family of
(A) straight lines
(B) parabolas
(C) Circles
(D) Ellipses

Showing the 12 most recent of 45 on this concept.

1. TG EAPCET 2021Set eng-2021-08-05-FN1 markMCQ
   Q.Let $(x, y) \in (\mathbb{R} \times \mathbb{R})$ and $\vec{a} = xi + 2j - k$, $\vec{b} = 6i - yj + 2k$ be two vectors. If $\left| \vec{a} \times \vec{b} \right|^2 + \left| \vec{a} \cdot \vec{b} \right|^2 = f(x)g(y)$ then $f(x) + g(y) - 46 = 0$ represents (A) a pair of lines (B) an ellipse (C) a hyperbola (D) a circle

   ›Reveal solutionSolution

   The identity $\left| \vec{a} \times \vec{b} \right|^2 + \left| \vec{a} \cdot \vec{b} \right|^2 = |\vec{a}|^2 |\vec{b}|^2$ lets us factor $f(x)g(y)$ as $(x^2+5)(y^2+40)$. Then $f(x)+g(y)-46=0$ becomes $x^2+y^2=1$, which is a circle.

   The core idea is a fundamental vector identity: for any two vectors $\vec{a}$ and $\vec{b}$, the sum of the squared magnitudes of their cross product and dot product equals the product of the squares of their magnitudes. That is,

   $|\vec{a} \times \vec{b}|^2 + |\vec{a} \cdot \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2.$

   This identity is a direct consequence of the relation $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}||\sin\theta|$ and $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$, so the left side becomes $|\vec{a}|^2|\vec{b}|^2(\sin^2\theta + \cos^2\theta) = |\vec{a}|^2|\vec{b}|^2$.

   Here, we are told that this sum equals $f(x)g(y)$, meaning it factors neatly into a product of a function of $x$ alone and a function of $y$ alone. Once we compute $|\vec{a}|^2$ and $|\vec{b}|^2$, we can identify $f(x)$ and $g(y)$. Then the equation $f(x)+g(y)-46=0$ simplifies to a familiar conic.

   1. Compute $|\vec{a}|^2$ and $|\vec{b}|^2$. $\vec{a} = x\hat{i} + 2\hat{j} - \hat{k}$, so

   $|\vec{a}|^2 = x^2 + 2^2 + (-1)^2 = x^2 + 4 + 1 = x^2 + 5.$

   $\vec{b} = 6\hat{i} - y\hat{j} + 2\hat{k}$, so

   $|\vec{b}|^2 = 6^2 + (-y)^2 + 2^2 = 36 + y^2 + 4 = y^2 + 40.$

   2. Apply the identity. Using $|\vec{a} \times \vec{b}|^2 + |\vec{a} \cdot \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2$, we get

   $|\vec{a} \times \vec{b}|^2 + |\vec{a} \cdot \vec{b}|^2 = (x^2+5)(y^2+40).$

   The problem states this equals $f(x)g(y)$. So we can take …

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2. TG EAPCET 2022Set eng-2022-07-19-FN1 markMCQ
   Q.If the distance from a variable point P to a fixed point $A(a,0)$ is equal to the perpendicular distance from P to the line $x + y = 0$ then the equation of the locus of P is (A) $x^2 + y^2 - 2xy - 4ax = 0$ (B) $x^2 + y^2 - 2xy - 4ax + 2a^2 = 0$ (C) $x^2 - 4ay + y^2 = 0$ (D) $(x-a)^2 + y^2 = 4axy$

   ›Reveal solutionSolution

   The locus of a point P whose distance from a fixed point A is equal to its perpendicular distance from a fixed line is a parabola. By equating these two distances and simplifying, we find the equation of the locus. The correct option is (B).

   Concept and Intuition

   The problem asks for the equation of the locus of a point P. A "locus" is simply the path traced by a point that moves according to a given condition. To find the equation of a locus, we typically:

   1. Assume the variable point P has coordinates $(x, y)$.
   2. Translate the given condition into an algebraic equation involving $x$ and $y$.
   3. Simplify this equation.

   In this specific problem, the condition is that the distance from P to a fixed point $A(a,0)$ is equal to the perpendicular distance from P to the line $x + y = 0$. This is the defining characteristic of a parabola: a parabola is the set of all points that are equidistant from a fixed point (the focus) and a fixed line (the directrix). Here, $A(a,0)$ is the focus and $x+y=0$ is the directrix.

   Our strategy will be to express both distances algebraically using the coordinates of P and the given point/line, then set them equal to each other and simplify.

   Step-by-step Derivation

   1. Define the variable point P:

      Let the coordinates of the variable point P be $(x, y)$.

   2. Identify the fixed point A:

      The fixed point A is given as $(a, 0)$.

   3. Calculate the distance PA:

      Using the distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$, which is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:

      $PA = \sqrt{(x-a)^2 + (y-0)^2}$

      $PA = \sqrt{(x-a)^2 + y^2}$

   4. Identify the fixed line:

      The fixed line is given by the equation $x + y = 0$.

   5. Calculate the perpendicular distance from P to the line:

      The formula for the perpendicular distance from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$ is $\frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.

      Here, the point is P$(x, y)$, and the line is $1x + 1y + 0 = 0$. So, $A=1$, $B=1$, $C=0$.

      Perpendicular distance $PL = \frac{|1 \cdot x + 1 \cdot y + 0|}{\sqrt{1^2 + 1^2}}$

      $PL = \frac{|x + y|}{\sqrt{2}}$

   6. Equate the distances:

      According to the problem statement, the distance from P to A is equal to the perpendicular distance from P to the line:

      $PA = PL$

      $\sqrt{(x-a)^2 + y^2} = \frac{|x + y|}{\sqrt{2}}$

   7. Square both sides to eliminate the square root and absolute value:

      Squaring both sides will remove the square root on the left and the absolute value on the right (since $(|k|)^2 = k^2$).

      $(\sqrt{(x-a)^2 + y^2})^2 = \left(\frac{|x + y|}{\sqrt{2}}\right)^2$

      $(x-a)^2 + y^2 = \frac{(x + y)^2}{2}$ …

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3. TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
   Q.A focus of an ellipse having eccentricity $\dfrac{1}{2}$ is at (0, 0) and a directrix is the line $x = 4$. Then the equation of one such ellipse is (A) $\dfrac{9x^2}{64} + \dfrac{3y^2}{16} = 1$ (B) $\dfrac{(2x+1)^2}{32} + \dfrac{y^2}{16} = 1$ (C) $\dfrac{(3x+4)^2}{64} + \dfrac{y^2}{32} = 1$ (D) $(3x+4)^2 + 12y^2 = 64$

   ›Reveal solutionSolution

   Using the focus-directrix property of an ellipse, we set the distance from any point to the focus equal to $e$ times the perpendicular distance to the directrix, then simplify to get the Cartesian equation. The correct ellipse is given by option (D).

   The key idea is that an ellipse is defined as the set of points whose distance from a fixed point (the focus) is a constant fraction $e$ (the eccentricity) of its perpendicular distance from a fixed line (the directrix). Here $e = \frac12$, focus at $(0,0)$, and directrix $x = 4$.

   Let’s work through it step by step.

   1. Set up the distance condition. For any point $(x, y)$ on the ellipse, the distance to the focus $(0,0)$ is $\sqrt{x^2 + y^2}$. The perpendicular distance from $(x, y)$ to the directrix $x = 4$ is $|x - 4|$. The focus-directrix relation gives:

   $\sqrt{x^2 + y^2} = e \cdot |x - 4| = \frac12 |x - 4|.$

   2. Square both sides to remove the square root and absolute value.

   $x^2 + y^2 = \frac14 (x - 4)^2.$

   3. Clear the fraction and expand. Multiply through by 4:

   $4x^2 + 4y^2 = (x - 4)^2 = x^2 - 8x + 16.$

   4. Bring all terms to one side.

   $4x^2 + 4y^2 - x^2 + 8x - 16 = 0 \quad\Rightarrow\quad 3x^2 + 8x + 4y^2 - 16 = 0.$

   5. Complete the square for the $x$-terms.

      Factor the $x$-terms: $3x^2 + 8x = 3\left(x^2 + \frac{8}{3}x\right)$.

      Complete the square inside: $x^2 + \frac{8}{3}x = \left(x + \frac{4}{3}\right)^2 - \frac{16}{9}$.

      So $3x^2 + 8x = 3\left(x + \frac{4}{3}\right)^2 - \frac{16}{3}$.

   6. Substitute back into the equation.

   $3\left(x + \frac{4}{3}\right)^2 - \frac{16}{3} + 4y^2 - 16 = 0.$

   Combine constants: $-\frac{16}{3} - 16 = -\frac{16}{3} - \frac{48}{3} = -\frac{64}{3}$.

   So:

   $3\left(x + \frac{4}{3}\right)^2 + 4y^2 = \frac{64}{3}.$

   7. Divide through to get the standard form. …

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4. TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQ
   Q.The locus of a point P which moves such that the sum of its distances from two perpendicular lines is equal to 1 is a (A) Square (B) Circle (C) Straight line (D) set of four parallel lines

   ›Reveal solutionSolution

   The problem reduces to $|x| + |y| = 1$ in a coordinate system aligned with the two perpendicular lines. This equation describes a square rotated by $45^\circ$ — a diamond shape — so the correct option is (A).

   The key idea is to choose the two perpendicular lines as the coordinate axes. Why? Because the distance from a point to a line is a simple absolute value when the line is an axis. The sum of distances then becomes $|x| + |y|$, and the condition $|x| + |y| = 1$ is a classic shape — a square whose sides are at $45^\circ$ to the axes.

   Let’s work through it.

   1. Set up the coordinate system.

      Take the two given perpendicular lines as the $x$-axis and $y$-axis. This is always possible by translation and rotation — the geometry doesn’t change. Let $P = (x, y)$.

   2. Write the distances.

      Distance from $P$ to the $x$-axis is $|y|$.

      Distance from $P$ to the $y$-axis is $|x|$.

      The given condition: $|x| + |y| = 1$.

   3. Interpret the equation $|x| + |y| = 1$.

      This is not a circle (that would be $x^2 + y^2 = 1$). It’s not a single straight line. Let’s see what it looks like in each quadrant.

      • In Quadrant I ($x \ge 0, y \ge 0$): $x + y = 1$ — a line segment from $(1,0)$ to $(0,1)$.
      • In Quadrant II ($x \le 0, y \ge 0$): $-x + y = 1$ — a line segment from $(-1,0)$ to $(0,1)$.
      • In Quadrant III ($x \le 0, y \le 0$): $-x - y = 1$ — a line segment from $(-1,0)$ to $(0,-1)$.
      • In Quadrant IV ($x \ge 0, y \le 0$): $x - y = 1$ — a line segment from $(1,0)$ to $(0,-1)$.

      These four line segments join to form a closed shape. The vertices are $(1,0)$, $(0,1)$, $(-1,0)$, $(0,-1)$.

   4. Identify the shape. …

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5. TG EAPCET 2022Set eng-2022-07-19-FN1 markMCQ
   Q.The equation of the normal to the curve $4x^2 + 9y^2 = 36$ at the point $P\left(\dfrac{7\pi}{4}\right)$ is (A) $2x - 3y - 6\sqrt{2} = 0$ (B) $2x + 3y = 0$ (C) $3\sqrt{2}x + 2\sqrt{2}y - 5 = 0$ (D) $3\sqrt{2}x - 2\sqrt{2}y - 13 = 0$

   ›Reveal solutionSolution

   The curve is an ellipse, and the point is given in parametric form. We find the slope of the tangent using derivatives, then the slope of the normal, and finally the equation of the normal. The correct option is (D).

   The curve $4x^2 + 9y^2 = 36$ is an ellipse. Dividing through by 36 gives $\frac{x^2}{9} + \frac{y^2}{4} = 1$, so the standard parametric form is $x = 3\cos\theta$, $y = 2\sin\theta$. The point is given as $P\left(\frac{7\pi}{4}\right)$, meaning $\theta = \frac{7\pi}{4}$.

   The key idea: the normal line is perpendicular to the tangent. So we first find the slope of the tangent at that point, then take its negative reciprocal.

   1. Find the coordinates of P.

      At $\theta = \frac{7\pi}{4}$,

      $\cos\frac{7\pi}{4} = \frac{1}{\sqrt{2}}$ and $\sin\frac{7\pi}{4} = -\frac{1}{\sqrt{2}}$.

      So $x = 3 \cdot \frac{1}{\sqrt{2}} = \frac{3}{\sqrt{2}}$, $y = 2 \cdot \left(-\frac{1}{\sqrt{2}}\right) = -\frac{2}{\sqrt{2}} = -\sqrt{2}$.

      Thus $P\left(\frac{3}{\sqrt{2}}, -\sqrt{2}\right)$.

   2. Find the slope of the tangent using implicit differentiation.

      Differentiate $4x^2 + 9y^2 = 36$ with respect to $x$:

      $8x + 18y \frac{dy}{dx} = 0$

      $\Rightarrow \frac{dy}{dx} = -\frac{8x}{18y} = -\frac{4x}{9y}$.

      At $P$:

      $\frac{dy}{dx} = -\frac{4 \cdot \frac{3}{\sqrt{2}}}{9 \cdot (-\sqrt{2})} = -\frac{12/\sqrt{2}}{-9\sqrt{2}} = -\frac{12/\sqrt{2}}{-9\sqrt{2}}$.

      Simplify: $-\frac{12/\sqrt{2}}{-9\sqrt{2}} = \frac{12/\sqrt{2}}{9\sqrt{2}} = \frac{12}{9 \cdot 2} = \frac{12}{18} = \frac{2}{3}$.

      So the slope of the tangent is $\frac{2}{3}$.

   3. Slope of the normal.

      The normal is perpendicular to the tangent, so its slope $m_n$ satisfies $m_n \cdot \frac{2}{3} = -1$, giving $m_n = -\frac{3}{2}$.

   4. Equation of the normal. …

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6. TG EAPCET 2021Set eng-2021-08-06-FN1 markMCQ
   Q.The ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 (b > a)$ and the parabola $y^2 = 8ax$ cut at right angles. If $e$ is the eccentricity of the ellipse, then $e^4 =$ (A) $\dfrac{1}{4}$ (B) $\dfrac{1}{16}$ (C) $\dfrac{1}{8}$ (D) $\dfrac{1}{64}$

   ›Reveal solutionSolution

   The condition for orthogonal intersection gives a relation between the slopes of the tangents at the common point. Solving that yields $e^4 = \frac{1}{16}$.

   The key idea is that when two curves cut at right angles, their tangents at the intersection point are perpendicular. That means the product of their slopes is $-1$. So we need to find a common point of the ellipse and the parabola, compute the slopes of the tangents there, set the product to $-1$, and then use the eccentricity formula for the ellipse.

   Let’s go step by step.

   1. Find the intersection point(s) The parabola is $y^2 = 8ax$. Substitute into the ellipse equation:

   $\frac{x^2}{a^2} + \frac{8ax}{b^2} = 1$

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

   $b^2 x^2 + 8a^3 x = a^2 b^2$

   This is a quadratic in $x$. But notice that the curves are symmetric about the $x$-axis, and we expect a single intersection in the first quadrant (since $b > a$, the ellipse is vertical). The obvious candidate is $x = a$ — let’s check:

   If $x = a$, then from the parabola $y^2 = 8a \cdot a = 8a^2$, so $y = \pm 2\sqrt{2} a$.

   Plug into the ellipse: $\frac{a^2}{a^2} + \frac{8a^2}{b^2} = 1 + \frac{8a^2}{b^2}$. For this to equal $1$, we need $\frac{8a^2}{b^2} = 0$, which is false. So $x = a$ is not the intersection.

   Instead, solve the quadratic properly. The equation is:

   $b^2 x^2 + 8a^3 x - a^2 b^2 = 0$

   The discriminant:

   $\Delta = (8a^3)^2 + 4 b^2 \cdot a^2 b^2 = 64 a^6 + 4 a^2 b^4$

   That’s messy. But we can use a smarter approach: since the curves cut at right angles, the point of intersection must satisfy both equations, and we can work with the slopes directly without fully solving for $x$ and $y$.

   2. Slope of tangent to the parabola For $y^2 = 8ax$, differentiate implicitly:

   $2y \frac{dy}{dx} = 8a \quad \Rightarrow \quad \frac{dy}{dx} = \frac{4a}{y}$

   So at any point $(x_1, y_1)$ on the parabola, the slope $m_1 = \frac{4a}{y_1}$.

   3. Slope of tangent to the ellipse For $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, differentiate:

   $\frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0 \quad \Rightarrow \quad \frac{dy}{dx} = -\frac{b^2 x}{a^2 y}$

   So at the same point $(x_1, y_1)$, the slope $m_2 = -\frac{b^2 x_1}{a^2 y_1}$.

   4. Orthogonality condition The tangents are perpendicular, so $m_1 \cdot m_2 = -1$:

   $\frac{4a}{y_1} \cdot \left( -\frac{b^2 x_1}{a^2 y_1} \right) = -1$

   Simplify:

   $-\frac{4a b^2 x_1}{a^2 y_1^2} = -1 \quad \Rightarrow \quad \frac{4 b^2 x_1}{a y_1^2} = 1$ …

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7. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
   Q.If $\theta$ is the angle between the curves $y^2 = 4x$ and $x^2 + y^2 = 5$ then $|\tan \theta| =$ (A) 5 (B) 4 (C) 3 (D) 2

   ›Reveal solutionSolution

   The angle between two curves is the angle between their tangents at the intersection point. For the curves $y^2 = 4x$ and $x^2 + y^2 = 5$, the intersection points are $(1, \pm 2)$. Using slopes of tangents at $(1,2)$, we get $|\tan \theta| = 3$.

   The angle between two curves at a point of intersection is defined as the angle between their tangents at that point. So the problem reduces to finding the slopes of the two curves at their common point(s), then using the formula for the angle between two lines.

   First, find where the curves meet. The parabola $y^2 = 4x$ and the circle $x^2 + y^2 = 5$ intersect when we substitute $y^2 = 4x$ into the circle's equation:

   $x^2 + 4x = 5 \quad \Rightarrow \quad x^2 + 4x - 5 = 0$

   This factors as $(x+5)(x-1)=0$, so $x = 1$ or $x = -5$. Since $y^2 = 4x$ requires $x \ge 0$, discard $x = -5$. Thus $x = 1$, and then $y^2 = 4$ gives $y = \pm 2$. The two intersection points are $(1, 2)$ and $(1, -2)$. By symmetry, the angle between the curves will be the same at both points — we'll work with $(1, 2)$.

   1. Slope of the parabola $y^2 = 4x$ at $(1,2)$

   Differentiate implicitly: $2y \frac{dy}{dx} = 4$, so $\frac{dy}{dx} = \frac{4}{2y} = \frac{2}{y}$. At $(1,2)$, the slope is $m_1 = \frac{2}{2} = 1$.

   2. Slope of the circle $x^2 + y^2 = 5$ at $(1,2)$

   Differentiate implicitly: $2x + 2y \frac{dy}{dx} = 0$, so $\frac{dy}{dx} = -\frac{x}{y}$. At $(1,2)$, the slope is $m_2 = -\frac{1}{2}$.

   3. Angle between the tangents

   If two lines have slopes $m_1$ and $m_2$, the acute angle $\theta$ between them satisfies

   $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$

   Here $m_1 = 1$ and $m_2 = -\frac{1}{2}$, so

   $m_1 - m_2 = 1 - \left(-\frac{1}{2}\right) = \frac{3}{2}$

   $1 + m_1 m_2 = 1 + (1)\left(-\frac{1}{2}\right) = 1 - \frac{1}{2} = \frac{1}{2}$

   Thus …

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8. TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
   Q.If A (1,1), B(-1,1) and C(-1,-1) are three points and a point P moves such that $PA^2 = PB^2 + PC^2$, then the equation of the locus of P is (A) $x^2 + y^2 - 6x - 2y + 2 = 0$ (B) $x^2 + y^2 + 6x + 2y + 2 = 0$ (C) $x^2 + y^2 + 6x - 2y + 2 = 0$ (D) $x^2 + y^2 + 6x + 2y - 2 = 0$

   ›Reveal solutionSolution

   The locus of a point P satisfying a given geometric condition is the path traced by P. By setting P as $(x, y)$ and using the distance formula, the condition $PA^2 = PB^2 + PC^2$ simplifies to the equation of a circle. The equation of the locus of P is $\boxed{x^2 + y^2 + 6x + 2y + 2 = 0}$.

   The problem asks for the equation of the locus of a point P that moves according to a specific geometric condition involving its distances from three fixed points A, B, and C. The "locus" of a point is simply the set of all possible positions that the point can occupy while satisfying the given condition. To find this equation, we represent the moving point P with general coordinates $(x, y)$ and then translate the given condition into an algebraic equation involving $x$ and $y$.

   The core idea is to use the distance formula. If we have two points $(x_1, y_1)$ and $(x_2, y_2)$, the distance $d$ between them is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. The problem uses squared distances ($PA^2$, $PB^2$, $PC^2$), which simplifies calculations by removing the square root.

   Here's how to derive the equation step-by-step:

   1. Define the coordinates of the points.

      Let the coordinates of the moving point P be $(x, y)$.

      The given fixed points are:

      A = $(1, 1)$

      B = $(-1, 1)$

      C = $(-1, -1)$

   2. Calculate the squared distances $PA^2$, $PB^2$, and $PC^2$.

      Using the distance formula squared, $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$:

      $PA^2 = (x - 1)^2 + (y - 1)^2$

      $PB^2 = (x - (-1))^2 + (y - 1)^2 = (x + 1)^2 + (y - 1)^2$

      $PC^2 = (x - (-1))^2 + (y - (-1))^2 = (x + 1)^2 + (y + 1)^2$

   3. Substitute these expressions into the given condition.

      The condition for point P is $PA^2 = PB^2 + PC^2$.

      Substituting the expressions from Step 2:

      $(x - 1)^2 + (y - 1)^2 = \left[ (x + 1)^2 + (y - 1)^2 \right] + \left[ (x + 1)^2 + (y + 1)^2 \right]$

   4. Expand and simplify the equation.

      Expand each squared term:

      $(x^2 - 2x + 1) + (y^2 - 2y + 1) = (x^2 + 2x + 1) + (y^2 - 2y + 1) + (x^2 + 2x + 1) + (y^2 + 2y + 1)$

      Combine like terms on the right side of the equation:

      $x^2 - 2x + y^2 - 2y + 2 = (x^2 + x^2) + (y^2 + y^2) + (2x + 2x) + (-2y + 2y) + (1 + 1 + 1 + 1)$

      $x^2 - 2x + y^2 - 2y + 2 = 2x^2 + 2y^2 + 4x + 4$ …

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9. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
   Q.If $4x + 2y + n = 0$ is a normal to the ellipse $\dfrac{x^2}{36} + \dfrac{y^2}{16} = 1$, then $n =$ (A) $\pm \dfrac{9}{4}$ (B) $\pm \dfrac{9}{\sqrt{10}}$ (C) $\pm \dfrac{5}{4}$ (D) $\pm 8$

   ›Reveal solutionSolution

   Applying the normal-line condition for an ellipse to $4x+2y+n=0$ gives $n^2=64$, so $n=\pm 8$.

   Normal condition. For the line $lx+my+n=0$ to be a normal to $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$,

   $\frac{a^2}{l^2} + \frac{b^2}{m^2} = \frac{(a^2-b^2)^2}{n^2}.$

   Substitute. Here $a^2=36,\ b^2=16$, and the line is $4x+2y+n=0$ so $l=4,\ m=2$:

   $\frac{36}{16} + \frac{16}{4} = \frac{(36-16)^2}{n^2}$ …

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10. TG EAPCET 2022Set eng-2022-07-19-FN1 markMCQ
    Q.If $\theta$ is the acute angle between the curves $x^2 + y^2 = 4$ and $y^2 = 3x$ then $\tan \theta =$ (A) $\dfrac{5}{\sqrt{3}}$ (B) $\dfrac{\sqrt{3}}{4}$ (C) $\dfrac{4}{\sqrt{3}}$ (D) $\dfrac{\sqrt{3}}{5}$

    ›Reveal solutionSolution

    To find the angle between two curves, we first find their intersection points, then calculate the slopes of their tangents at these points, and finally use the formula for the angle between two lines. The tangent of the acute angle is $\boxed{\dfrac{5}{\sqrt{3}}}$.

    When we talk about the angle between two curves, we are actually referring to the angle between their tangent lines at a point of intersection. The core idea is that locally, near an intersection point, the curves can be approximated by their tangents. Therefore, the problem boils down to:

    1. Finding the point(s) where the curves intersect.
    2. Calculating the slope of the tangent to each curve at an intersection point.
    3. Using the formula for the angle between two lines given their slopes.

    Let's apply this step-by-step.

    1. Find the points of intersection of the two curves.

       The given curves are:

       Curve 1: $x^2 + y^2 = 4$ (a circle)

       Curve 2: $y^2 = 3x$ (a parabola)

       Substitute $y^2 = 3x$ from Curve 2 into Curve 1:

       $x^2 + (3x) = 4$

       $x^2 + 3x - 4 = 0$

       This is a quadratic equation in $x$. We can factor it:

       $(x+4)(x-1) = 0$

       This gives two possible values for $x$: $x = -4$ or $x = 1$.

       Now, we check these $x$ values with Curve 2, $y^2 = 3x$.

       If $x = -4$, then $y^2 = 3(-4) = -12$. Since $y^2$ cannot be negative for real $y$, $x = -4$ is not a valid x-coordinate for an intersection point.

       If $x = 1$, then $y^2 = 3(1) = 3$. This gives $y = \pm \sqrt{3}$.

       So, the curves intersect at two points: $(1, \sqrt{3})$ and $(1, -\sqrt{3})$. Due to the symmetry of the curves, the angle between them will be the same at both intersection points. We can choose either point; let's use $(1, \sqrt{3})$.

    2. Calculate the slopes of the tangents to each curve at the intersection point $(1, \sqrt{3})$.

       We find the derivative $\frac{dy}{dx}$ for each curve, which represents the slope of the tangent.

       • For Curve 1: $x^2 + y^2 = 4$

         Differentiate implicitly with respect to $x$:

         $\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(4)$

         $2x + 2y \frac{dy}{dx} = 0$

         $2y \frac{dy}{dx} = -2x$

         $\frac{dy}{dx} = -\frac{x}{y}$

         At the point $(1, \sqrt{3})$, the slope $m_1$ is:

         $m_1 = -\frac{1}{\sqrt{3}}$

       • For Curve 2: $y^2 = 3x$

         Differentiate implicitly with respect to $x$:

         $\frac{d}{dx}(y^2) = \frac{d}{dx}(3x)$

         $2y \frac{dy}{dx} = 3$

         $\frac{dy}{dx} = \frac{3}{2y}$

         At the point $(1, \sqrt{3})$, the slope $m_2$ is:

         $m_2 = \frac{3}{2\sqrt{3}}$

         We can simplify this by rationalizing the denominator, but it's not strictly necessary for the next step: $m_2 = \frac{3\sqrt{3}}{2 \cdot 3} = \frac{\sqrt{3}}{2}$.

    3. Use the formula for the angle between two lines.

       If $\theta$ is the acute angle between two lines with slopes $m_1$ and $m_2$, then: …

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11. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
    Q.Area of the quadrilateral formed by the common tangents drawn to the circle $x^2 + y^2 = 16$ and the ellipse $7x^2 + 25y^2 = 175$ is (A) $64$ (B) $32$ (C) $5\sqrt{2}$ (D) $16\sqrt{2}$

    ›Reveal solutionSolution

    The four common tangents to the circle and ellipse form a square whose diagonals lie along the coordinate axes. Finding the tangent lines and their intersection points gives an area of $64$ — option (A).

    The problem asks for the area of the quadrilateral formed by the common tangents to a circle and an ellipse, both centred at the origin. Because both curves are centred at the origin and symmetric about both axes, the common tangents come in symmetric pairs, and the quadrilateral they enclose is a square rotated $45^\circ$ relative to the axes.

    1. Rewrite the equations in standard form. The circle is $x^2 + y^2 = 16$, so its centre is $(0,0)$ and radius $r = 4$. The ellipse is $7x^2 + 25y^2 = 175$. Divide through by $175$:

    $\frac{x^2}{25} + \frac{y^2}{7} = 1$

    So $a^2 = 25$, $b^2 = 7$, with $a > b$, also centred at the origin.

    2. Set up the tangency condition for the circle. Let a tangent line be $y = mx + c$. For it to be tangent to the circle $x^2 + y^2 = 16$, the perpendicular distance from the centre to the line must equal the radius:

    $\frac{|c|}{\sqrt{1+m^2}} = 4 \quad \Rightarrow \quad c^2 = 16(1+m^2)$

    3. Apply the tangency condition for the ellipse. For the ellipse $\frac{x^2}{25} + \frac{y^2}{7} = 1$, the condition that $y = mx + c$ is tangent is:

    $c^2 = a^2 m^2 + b^2 = 25m^2 + 7$

    4. Equate the two expressions for $c^2$.

    $16 + 16m^2 = 25m^2 + 7$

    $16 - 7 = 25m^2 - 16m^2$

    $9 = 9m^2 \quad \Rightarrow \quad m^2 = 1 \quad \Rightarrow \quad m = \pm 1$

    5. Find the corresponding $c$ values. Using $c^2 = 16(1+m^2) = 16(2) = 32$, so $c = \pm 4\sqrt{2}$. Thus the four common tangents are:

    $y = x + 4\sqrt{2}, \quad y = x - 4\sqrt{2}, \quad y = -x + 4\sqrt{2}, \quad y = -x - 4\sqrt{2}$

    6. Find the vertices of the quadrilateral they enclose. Solving the four lines pairwise for their intersections:
       • $y=x+4\sqrt2$ and $y=-x+4\sqrt2$ intersect at $(0, 4\sqrt{2})$.
       • $y=x+4\sqrt2$ and $y=-x-4\sqrt2$ intersect at $(-4\sqrt{2}, 0)$.
       • $y=x-4\sqrt2$ and $y=-x+4\sqrt2$ intersect at $(4\sqrt{2}, 0)$. …

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12. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
    Q.The equation of the line perpendicular to the radical axis of two circles $x^2 + y^2 - 5x + 6y + 12 = 0$, $x^2 + y^2 + 6x - 4y - 14 = 0$ and passing through $(1, 1)$ is (A) $2x + 3y - 5 = 0$ (B) $x + y - 2 = 0$ (C) $10x + 11y - 21 = 0$ (D) $11x + 10y - 21 = 0$

    ›Reveal solutionSolution

    The radical axis of two circles is the line obtained by subtracting their equations. The required line is perpendicular to this radical axis and passes through $(1,1)$. The answer is option (D) $11x + 10y - 21 = 0$.

    The radical axis is a beautiful concept: for two circles, it is the set of points having equal power with respect to both circles. Algebraically, it’s simply the line you get by subtracting one circle equation from the other. Once you have that line, the problem becomes a straightforward geometry of perpendicular lines.

    1. Find the radical axis.

       Write the two circles:

       $S_1: x^2 + y^2 - 5x + 6y + 12 = 0$

       $S_2: x^2 + y^2 + 6x - 4y - 14 = 0$

       Subtract $S_1$ from $S_2$ (or vice versa — the sign only flips the line, not its direction):

       $(S_2 - S_1): (x^2 + y^2 + 6x - 4y - 14) - (x^2 + y^2 - 5x + 6y + 12) = 0$

       The $x^2$ and $y^2$ cancel, leaving:

       $6x - 4y - 14 + 5x - 6y - 12 = 0$

       Simplify: $11x - 10y - 26 = 0$

       So the radical axis is $11x - 10y - 26 = 0$.

    2. Find the slope of the radical axis.

       Rewrite $11x - 10y - 26 = 0$ as $y = \frac{11}{10}x - \frac{26}{10}$.

       Its slope is $m_1 = \frac{11}{10}$.

    3. Slope of the perpendicular line.

       For perpendicular lines, $m_1 \cdot m_2 = -1$. …

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