Q.Form the differential equation by eliminating $A$ and $B$ in $Ax^2+By^2=1$.

Showing the 12 most recent of 20 on this concept.

1. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.The differential equation of the family of all circles of radius ‘a’ is (A) $y_2^2 + 1 = y_1^2 + a^2$ (B) $1 + y_1^2 = y_2^2 + a^2$ (C) $y_1 y_2 + (1 + y_1^2) = a$ (D) $(1 + y_1^2)^3 = a^2 y_2^2$

   ›Reveal solutionSolution

   The key idea is to start with the general equation of a circle of fixed radius $a$, eliminate the two arbitrary constants (centre coordinates) by differentiating twice, and then simplify to obtain the differential equation. The correct result is $(1 + y_1^2)^3 = a^2 y_2^2$, which matches option (D).

   The problem asks for the differential equation that represents all circles of a given radius $a$, regardless of where their centre lies. That means the family has two free parameters — the $x$ and $y$ coordinates of the centre. To eliminate them, we need to differentiate the circle’s equation twice, because each differentiation removes one constant. The final relation between $y$, $y_1$ (first derivative), and $y_2$ (second derivative) must be free of the centre coordinates.

   Let’s work through it.

   1. Write the general equation of a circle of radius $a$. Let the centre be at $(h, k)$. Then

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

   Here $h$ and $k$ are the arbitrary constants we need to eliminate.

   2. Differentiate once with respect to $x$. Using the chain rule:

   $2(x - h) + 2(y - k) y_1 = 0,$

   where $y_1 = \frac{dy}{dx}$. Divide through by 2:

   $(x - h) + (y - k) y_1 = 0. \qquad(1)$

   3. Differentiate a second time. Differentiate (1) with respect to $x$:

   $1 + (y - k) y_2 + y_1 \cdot y_1 = 0,$

   because the derivative of $(y - k) y_1$ is $(y - k) y_2 + y_1^2$. So

   $1 + (y - k) y_2 + y_1^2 = 0. \qquad(2)$

   4. Eliminate $(y - k)$ from (1) and (2). From (1), we have $(x - h) = - (y - k) y_1$. But we don’t need $x - h$ directly — we need to get rid of $(y - k)$. From (2):

   $(y - k) y_2 = - (1 + y_1^2).$

   So

   $y - k = -\frac{1 + y_1^2}{y_2}, \quad \text{provided } y_2 \neq 0.$

   5. Now eliminate $x - h$ using (1). From (1): $x - h = - (y - k) y_1$. Substitute the expression for $y - k$:

   $x - h = - \left( -\frac{1 + y_1^2}{y_2} \right) y_1 = \frac{y_1 (1 + y_1^2)}{y_2}.$

   6. Plug both $x - h$ and $y - k$ back into the original circle equation. The original equation is $(x - h)^2 + (y - k)^2 = a^2$. Substituting:

   $\left( \frac{y_1 (1 + y_1^2)}{y_2} \right)^2 + \left( -\frac{1 + y_1^2}{y_2} \right)^2 = a^2.$

   Factor $(1 + y_1^2)^2 / y_2^2$ out of both terms:

   $\frac{(1 + y_1^2)^2}{y_2^2} \left( y_1^2 + 1 \right) = a^2.$ …

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2. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.The differential equation of the family of all circles of radius ‘a’ is (A) $y_1 y_2 + (1 + y_1^2) = a$ (B) $(1 + y_1^2)^3 = a^2 y_2^2$ (C) $1 + y_1^2 = y_2^2 + a^2$ (D) $y_2^2 + 1 = y_1^2 + a^2$

   ›Reveal solutionSolution

   The family of all circles of fixed radius $a$ has a differential equation that eliminates the two arbitrary parameters (center coordinates). The correct equation is $(1 + y_1^2)^3 = a^2 y_2^2$, which is option (B).

   We start with the general equation of a circle of radius $a$:

   $(x - h)^2 + (y - k)^2 = a^2$

   Here $h$ and $k$ are the coordinates of the center — two arbitrary constants. To get a differential equation that describes all such circles (no matter where they are placed), we must eliminate $h$ and $k$ by differentiating.

   Why this approach works:

   Each differentiation reduces the number of arbitrary constants. Two constants require two derivatives. The resulting relation between $y, y_1, y_2$ (where $y_1 = dy/dx$, $y_2 = d^2y/dx^2$) will be free of $h$ and $k$ and will involve only $a$.

   ---

   1. First derivative Differentiate the circle equation implicitly with respect to $x$:

   $2(x - h) + 2(y - k) y_1 = 0$

   Divide by 2:

   $(x - h) + (y - k) y_1 = 0 \quad \text{(1)}$

   This gives a linear relation between $x - h$ and $y - k$.

   2. Second derivative Differentiate (1) again with respect to $x$:

   $1 + (y - k) y_2 + y_1^2 = 0$

   (Remember: derivative of $(y - k) y_1$ is $y_1 \cdot y_1 + (y - k) y_2 = y_1^2 + (y - k) y_2$.)

   So:

   $1 + y_1^2 + (y - k) y_2 = 0 \quad \text{(2)}$

   3. Eliminate $y - k$ From (2):

   $y - k = -\frac{1 + y_1^2}{y_2}$

   Substitute into (1):

   $x - h + \left(-\frac{1 + y_1^2}{y_2}\right) y_1 = 0$

   So:

   $x - h = \frac{y_1(1 + y_1^2)}{y_2}$

   4. Use the original circle equation Plug $x - h$ and $y - k$ into $(x - h)^2 + (y - k)^2 = a^2$: $$$$

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3. TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
   Q.If $m$ and $n$ are respectively the order and the degree of the differential equation representing the family of curves $y^2 - 5ax - 5a^{\frac{3}{2}} = 0$ ($a > 0$ is a parameter), then the value of $m - n$ is (A) $1$ (B) $-1$ (C) $2$ (D) $-2$

   ›Reveal solutionSolution

   The key idea is to eliminate the parameter $a$ from the given family to obtain the differential equation, then identify its order $m$ and degree $n$. The value of $m - n$ is $\boxed{-1}$.

   The problem gives a family of curves with a single parameter $a$. To find the differential equation that represents this family, we must eliminate $a$ by differentiating and then combining the equations. The order of the resulting differential equation is the highest derivative present, and the degree is the power of the highest derivative after the equation is made polynomial in derivatives.

   Let’s work through it step by step.

   1. Write the given family and differentiate once. The family is

   $y^2 - 5ax - 5a^{3/2} = 0, \quad a > 0.$

   Differentiate both sides with respect to $x$:

   $2y \frac{dy}{dx} - 5a = 0.$

   So

   $a = \frac{2y}{5} \frac{dy}{dx}.$

   2. Substitute $a$ back into the original equation. Replace $a$ in $y^2 - 5ax - 5a^{3/2} = 0$:

   $y^2 - 5\left(\frac{2y}{5} \frac{dy}{dx}\right)x - 5\left(\frac{2y}{5} \frac{dy}{dx}\right)^{3/2} = 0.$

   Simplify the second term:

   $y^2 - 2xy \frac{dy}{dx} - 5\left(\frac{2y}{5} \frac{dy}{dx}\right)^{3/2} = 0.$

   3. Isolate the term with the fractional exponent. Move the other terms to the other side:

   $5\left(\frac{2y}{5} \frac{dy}{dx}\right)^{3/2} = y^2 - 2xy \frac{dy}{dx}.$

   4. Square both sides to remove the $3/2$ exponent. This is necessary to get a polynomial in derivatives (so we can read the degree). Squaring:

   $25 \left(\frac{2y}{5} \frac{dy}{dx}\right)^3 = \left(y^2 - 2xy \frac{dy}{dx}\right)^2.$

   Simplify the left side:

   $25 \cdot \frac{8y^3}{125} \left(\frac{dy}{dx}\right)^3 = \frac{8y^3}{5} \left(\frac{dy}{dx}\right)^3.$ …

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4. TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
   Q.If the order and degree of the differential equation corresponding to the family of curves $y^2 = 4a(x+a)$ ($a$ is parameter) are $m$ and $n$ respectively, then $m+n^2=$ (A) 3 (B) 4 (C) 5 (D) 2

   ›Reveal solutionSolution

   The differential equation for the family $y^2 = 4a(x+a)$ has order $m = 1$ and degree $n = 2$, so $m + n^2 = 1 + 4 = 5$. The correct option is (C).

   ---

   We start with the family of curves given by

   $y^2 = 4a(x + a),$

   where $a$ is a parameter. To find the differential equation, we eliminate $a$ by differentiating and then combining the equations.

   Concept & Intuition:

   A family of curves with one parameter yields a first-order differential equation. The order is the highest derivative present; the degree is the power of that highest derivative after the equation is made polynomial in derivatives. Here, because the parameter appears squared in the constant term, the elimination will produce a squared first derivative, giving degree 2.

   ---

   Step-by-step solution:

   1. Differentiate once with respect to $x$:

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

   where $y' = \frac{dy}{dx}$.

   2. Express $a$ in terms of $y$ and $y'$:

   $a = \frac{y y'}{2}.$

   3. Substitute back into the original equation to eliminate $a$:

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

   Simplify step by step:

   $y^2 = 2 y y' \left( x + \frac{y y'}{2} \right) = 2xy y' + y^2 (y')^2.$

   4. Rearrange to standard form:

   $y^2 = 2xy y' + y^2 (y')^2 \quad \Rightarrow \quad 0 = 2xy y' + y^2 (y')^2 - y^2.$

   Factor $y$ (assuming $y \neq 0$ for non-degenerate curves):

   $y \left[ 2x y' + y (y')^2 - y \right] = 0.$

   Since $y = 0$ gives only a trivial case, the differential equation is

   $2x y' + y (y')^2 - y = 0.$

   5. Determine order and degree: …

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5. TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
   Q.The differential equation of the family of circles with fixed radius $r$ units and centre on the line $y = 3$, is (A) $1 + \left( \dfrac{dy}{dx} \right)^2 = \dfrac{r^2}{(y-3)^2}$ (B) $1 + \left( \dfrac{dy}{dx} \right)^2 = \dfrac{r^2}{y-3}$ (C) $\left( \dfrac{dy}{dx} \right)^2 = \dfrac{-r^2}{(y-3)^2}$ (D) $\left( \dfrac{dy}{dx} \right)^2 = \dfrac{r^2}{y-3}$

   ›Reveal solutionSolution

   The family of circles has centre $(h,3)$ and fixed radius $r$; eliminating the parameter $h$ by differentiating the circle equation and substituting back yields the differential equation $1 + (dy/dx)^2 = r^2/(y-3)^2$, which matches option (A).

   Concept & Intuition

   We have a family of curves (circles) that share a fixed radius $r$ and whose centres all lie on the horizontal line $y=3$. The only thing that varies from one circle to another is the $x$-coordinate of the centre, say $h$. To find the differential equation of the family, we need an equation that relates $x$, $y$, and $dy/dx$ without the parameter $h$. The standard approach: write the general equation of a circle in the family, differentiate it with respect to $x$, and then eliminate $h$ using the original equation.

   ---

   Step-by-step solution

   1. Write the equation of a typical circle A circle with centre $(h, 3)$ and radius $r$ has equation

   $(x - h)^2 + (y - 3)^2 = r^2.$

   Here $h$ is the parameter that distinguishes one circle from another.

   2. Differentiate implicitly with respect to $x$ Differentiating both sides:

   $2(x - h) + 2(y - 3) \frac{dy}{dx} = 0.$

   Divide through by 2:

   $(x - h) + (y - 3) \frac{dy}{dx} = 0.$

   This gives a relation between $x$, $y$, $h$, and $dy/dx$.

   3. Solve for $x - h$ From the differentiated equation:

   $x - h = - (y - 3) \frac{dy}{dx}.$

   4. Eliminate $h$ using the original circle equation Substitute $x - h$ from step 3 into the original circle equation:

   $\left[ - (y - 3) \frac{dy}{dx} \right]^2 + (y - 3)^2 = r^2.$

   Simplify:

   $(y - 3)^2 \left( \frac{dy}{dx} \right)^2 + (y - 3)^2 = r^2.$

   5. Factor and rearrange Factor $(y - 3)^2$: $$$$

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6. TG EAPCET 2021Set eng-2021-08-05-FN1 markMCQ
   Q.The solution of $\dfrac{dy}{dx} = \sqrt{1-y^2},\ y(0)=1,$ is (A) $\sin^{-1} y = x - \sin^{-1}(1)$ (B) $\sin^{-1} y = x + \sin^{-1}(1)$ (C) $\cos^{-1} y = x + \cos^{-1}(1)$ (D) $\sin^{-1} y + x = \sin^{-1}(1)$

   ›Reveal solutionSolution

   The key idea is to separate variables and integrate, then apply the initial condition $y(0)=1$ to fix the constant. The correct solution is $\sin^{-1} y = x + \sin^{-1}(1)$, which corresponds to option (B).

   The differential equation $\frac{dy}{dx} = \sqrt{1-y^2}$ is a first-order separable equation. The presence of $\sqrt{1-y^2}$ immediately suggests an inverse trigonometric function — either $\sin^{-1} y$ or $\cos^{-1} y$ — because the derivative of $\sin^{-1} y$ is $\frac{1}{\sqrt{1-y^2}}$, and here we have $\sqrt{1-y^2}$ in the denominator's place after separation. That’s the conceptual hook: we’re looking for a function whose derivative gives back the reciprocal of what we have.

   Let’s work through it.

   1. Separate the variables. Write

   $\frac{dy}{\sqrt{1-y^2}} = dx.$

   This is valid as long as $y \neq \pm 1$, but we’ll handle the initial condition separately.

   2. Integrate both sides. The left-hand side integrates to $\sin^{-1} y$ (since $\frac{d}{dy} \sin^{-1} y = \frac{1}{\sqrt{1-y^2}}$). So

   $\int \frac{dy}{\sqrt{1-y^2}} = \int dx \quad\Rightarrow\quad \sin^{-1} y = x + C,$

   where $C$ is the constant of integration.

   3. Apply the initial condition $y(0)=1$. Substitute $x=0$ and $y=1$:

   $\sin^{-1}(1) = 0 + C \quad\Rightarrow\quad C = \sin^{-1}(1).$

   So the particular solution is

   $\sin^{-1} y = x + \sin^{-1}(1).$

   4. Check against the options. This matches option (B) exactly. Option (A) has a minus sign, (C) uses $\cos^{-1}$ with a different constant, and (D) moves $x$ to the other side incorrectly. …

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7. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.If $y = \sin ax + \cos bx$ then $y'' + b^2 y =$ (A) $(b^2 - a^2) \sin ax$ (B) $(b^2 - a^2) \cos bx$ (C) $(a^2 - b^2) \tan ax$ (D) $(b^2 - a^2) \cot bx$

   ›Reveal solutionSolution

   The key idea is to compute the second derivative of $y = \sin ax + \cos bx$ and then substitute into $y'' + b^2 y$. The cross terms cancel, leaving only $(b^2 - a^2)\sin ax$, so the correct option is (A).

   We start with the function

   $y = \sin(ax) + \cos(bx).$

   The problem asks for $y'' + b^2 y$. The natural approach is to differentiate twice and then combine terms. Notice that the second derivative of $\sin(ax)$ will bring down a factor of $-a^2$, and the second derivative of $\cos(bx)$ will bring down a factor of $-b^2$. When we add $b^2 y$, the $\cos(bx)$ part will cancel, leaving only a term involving $\sin(ax)$. This is the core insight.

   Let’s work through it step by step.

   1. First derivative Differentiate each term:

   $y' = a\cos(ax) - b\sin(bx).$

   (Derivative of $\sin(ax)$ is $a\cos(ax)$; derivative of $\cos(bx)$ is $-b\sin(bx)$.)

   2. Second derivative Differentiate again:

   $y'' = -a^2\sin(ax) - b^2\cos(bx).$

   (Derivative of $a\cos(ax)$ is $-a^2\sin(ax)$; derivative of $-b\sin(bx)$ is $-b^2\cos(bx)$.)

   3. Form the expression $y'' + b^2 y$ Substitute $y''$ and $y$:

   $y'' + b^2 y = \bigl[-a^2\sin(ax) - b^2\cos(bx)\bigr] + b^2\bigl[\sin(ax) + \cos(bx)\bigr].$

   4. Simplify Group the $\sin(ax)$ terms and the $\cos(bx)$ terms: $$$$

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8. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.The general solution of the differential equation $(3x^2-2xy)dy+(y^2-2xy)dx=0$ is (A) $x^2-xy=cy^2$ (B) $y^2-xy=cx^3$ (C) $xy-x^2=cy^3$ (D) $xy-y^2=cx^3$

   ›Reveal solutionSolution

   This is a homogeneous differential equation. Substituting $y = vx$ reduces it to a separable form, leading to the solution $xy - x^2 = c\,y^3$, which corresponds to option (C).

   We start by recognizing the structure: the equation

   $(3x^2 - 2xy)\,dy + (y^2 - 2xy)\,dx = 0$

   has every term of total degree 2 (e.g., $3x^2$, $-2xy$, $y^2$ are all degree 2). That is the hallmark of a homogeneous differential equation — one where $M(x,y)$ and $N(x,y)$ are homogeneous functions of the same degree. For such equations, the substitution $y = vx$ turns it into a separable equation in $v$ and $x$.

   Let’s work through it step by step.

   1. Rewrite in standard form We have $M(x,y)\,dx + N(x,y)\,dy = 0$ with

   $M(x,y) = y^2 - 2xy, \qquad N(x,y) = 3x^2 - 2xy,$

   both homogeneous of degree 2.

   2. Substitute $y = vx$ Then $dy = v\,dx + x\,dv$, and

   $M = (vx)^2 - 2x(vx) = x^2(v^2 - 2v), \qquad N = 3x^2 - 2x(vx) = x^2(3 - 2v).$

   The equation becomes

   $x^2(v^2 - 2v)\,dx + x^2(3 - 2v)(v\,dx + x\,dv) = 0.$

   3. Divide through by $x^2$ (valid for $x \neq 0$)

   $(v^2 - 2v)\,dx + (3 - 2v)(v\,dx + x\,dv) = 0.$

   4. Collect the $dx$ and $dv$ terms

   $\big[(v^2 - 2v) + v(3 - 2v)\big]\,dx + (3 - 2v)x\,dv = 0.$

   Simplify the bracket:

   $(v^2 - 2v) + (3v - 2v^2) = -v^2 + v = v(1 - v).$

   So

   $v(1 - v)\,dx + (3 - 2v)x\,dv = 0.$

   5. Separate variables

   $\frac{dx}{x} = -\frac{3 - 2v}{v(1 - v)}\,dv.$

   6. Partial fractions Write $\dfrac{3 - 2v}{v(1 - v)} = \dfrac{A}{v} + \dfrac{B}{1 - v}$, so $3 - 2v = A(1 - v) + Bv$.
      • Set $v = 0$: $3 = A \Rightarrow A = 3$.
      • Set $v = 1$: $1 = B \Rightarrow B = 1$. Hence

   $\frac{3 - 2v}{v(1 - v)} = \frac{3}{v} + \frac{1}{1 - v}.$

   7. Integrate both sides $$$$

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9. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.The substitution required to reduce the differential equation $t^2 dx + (x^2 - tx + t^2) dt = 0$ to a differential equation which can be solved by variables separable method is (A) $t = Vx$ (B) $ax + bt = Z$ (C) $V = tx^2$ (D) $x = tV^2$

   ›Reveal solutionSolution

   The given differential equation is homogeneous in $x$ and $t$ (each term has total degree 2), so the standard substitution $x = V t$ (or equivalently $t = V x$) reduces it to a separable equation. The correct choice is (A).

   We are given:

   $t^2\,dx + (x^2 - tx + t^2)\,dt = 0$

   Concept & Intuition

   A differential equation of the form $M(x,t)\,dx + N(x,t)\,dt = 0$ is homogeneous if $M$ and $N$ are homogeneous functions of the same degree. Here, each term in $t^2$, $x^2$, and $tx$ is degree 2. For such equations, the substitution $x = V t$ (or $t = V x$) makes the equation separable. Why? Because dividing numerator and denominator by the highest power of $t$ (or $x$) turns the equation into one involving only the ratio $x/t$ (or $t/x$), which then separates.

   Let’s work through it step by step.

   1. Check homogeneity Rewrite the equation as:

   $t^2\,dx + (x^2 - t x + t^2)\,dt = 0$

   The coefficient of $dx$ is $t^2$ (degree 2 in $t$ and $x$ if we treat $t$ and $x$ as variables). The coefficient of $dt$ is $x^2 - t x + t^2$: each term is degree 2. So the equation is homogeneous of degree 2.

   2. Choose the substitution

      For a homogeneous equation, the standard substitution is $x = V t$ (or $t = V x$). This replaces the two variables with one variable $V$ and the independent variable $t$ (or $x$). The option (A) says $t = Vx$, which is the same idea but swapping roles. Let’s test $x = V t$ first (the more common form), then see why (A) works.

   3. Apply $x = V t$

      Let $x = V t$, so $dx = V\,dt + t\,dV$. Substitute into the equation:

   $t^2 (V\,dt + t\,dV) + \big((V t)^2 - t(V t) + t^2\big)\,dt = 0$

   Simplify:

   $t^2 V\,dt + t^3\,dV + (V^2 t^2 - V t^2 + t^2)\,dt = 0$

   Factor $t^2$ from the $dt$ terms:

   $t^2 V\,dt + t^3\,dV + t^2(V^2 - V + 1)\,dt = 0$

   Combine the $dt$ terms:

   $t^2\big(V + V^2 - V + 1\big)dt + t^3\,dV = 0$

   Notice $V - V$ cancels, leaving:

   $t^2(V^2 + 1)\,dt + t^3\,dV = 0$

   4. Separate variables Divide through by $t^3 (V^2+1)$ (assuming $t \neq 0$): $$$$

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10. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
    Q.The general solution of the differential equation $\dfrac{dy}{dx} + (\sec x \csc x) y = \cos^2 x$ is (A) $y \sec^2 x = \tan x + c$ (B) $y \tan x = \sin x \cos x + c$ (C) $y \sec^2 x = \sin^2 x + c$ (D) $2y \tan x = \sin^2 x + c$

    ›Reveal solutionSolution

    $2y\tan x = \sin^2 x + c$ — option (D).

    This is a linear first-order ODE $\dfrac{dy}{dx} + P(x)\,y = Q(x)$ with $P = \sec x\csc x = \dfrac{1}{\sin x\cos x}$ and $Q = \cos^2 x$.

    Integrating factor:

    $\int \frac{dx}{\sin x\cos x} = \int \frac{\sec^2 x}{\tan x}\,dx = \log|\tan x| \;\Rightarrow\; \text{IF} = \tan x.$

    Multiply through and integrate: …

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11. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
    Q.The general solution of the differential equation $\dfrac{dy}{dx} + (\sec x \csc x) y = \cos^2 x$ is (A) $y \sec^2 x = \sin^2 x + c$ (B) $y \sec^2 x = \tan x + c$ (C) $y \tan x = \sin x \cos x + c$ (D) $2y \tan x = \sin^2 x + c$

    ›Reveal solutionSolution

    A linear ODE with integrating factor $\tan x$: it integrates to $y\tan x=\tfrac12\sin^2 x+c$, i.e. $2y\tan x=\sin^2 x+c$. Answer: (D).

    Linear form. $\dfrac{dy}{dx}+P(x)y=Q(x)$ with $P=\sec x\csc x=\dfrac{1}{\sin x\cos x}$ and $Q=\cos^2 x$.

    Integrating factor.

    $\int P\,dx=\int\frac{dx}{\sin x\cos x}=\int\frac{2\,dx}{\sin 2x}=\int 2\csc 2x\,dx=\log|\tan x|,$

    so $\mu=e^{\log|\tan x|}=\tan x$.

    Multiply through. The coefficient of $y$ becomes $\tan x\cdot\sec x\csc x=\sec^2 x$, so the left side is an exact derivative: …

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12. TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
    Q.The equation $x^2 - y^2 + ax + b = 0$ represents a pair of lines for the ordered pair $(a, b) =$ (A) $(2, 6)$ (B) $(3, 4)$ (C) $(4, 8)$ (D) $(6, 9)$

    ›Reveal solutionSolution

    $x^2 - y^2 + ax + b = 0$ represents a pair of lines only if it factors as $(x-y+p)(x+y+q)=0$. The missing $y$-term forces $p=q$, giving $a = 2p$ and $b = p^2$, i.e. $a^2 = 4b$. Only $(a,b)=(6,9)$ satisfies this — option (D).

    The key idea: a second-degree equation in $x$ and $y$ represents a pair of straight lines exactly when it factors into two linear factors.

    Why this approach works.

    The quadratic part factors as $x^2 - y^2 = (x-y)(x+y)$, so the full expression should factor as $(x - y + p)(x + y + q) = 0$. Matching coefficients turns the condition into a simple relation between $a$ and $b$.

    Step-by-step solution.

    1. Assume the factored form.

    $(x - y + p)(x + y + q) = 0.$

    2. Expand.

    $(x - y + p)(x + y + q) = x^2 - y^2 + (p+q)x + (p - q)y + pq.$

    3. Match with $x^2 - y^2 + ax + b = 0$. There is no $y$-term, so its coefficient must vanish:

    $p - q = 0 \quad\Rightarrow\quad p = q.$

    4. Relate $a$ and $b$. With $p=q$: the coefficient of $x$ gives $a = 2p$, and the constant gives $b = p^2$. Eliminating $p$, $$b = \left(\frac{a}{2}\right)^2 \quad\Rightarrow\quad a^2 = 4b. …$$

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