Question

Consider a curve y = f(x) in xy plane. The curve passes through origin and has the property that a segment of tangent drawn at any point P(x, f(x)) and the line y = 3 gets bisected by the line x + y = 1, then the equation of the curve is $(y-p)^2 = q(r-x-y)$, where p, q, r are integers. Then $p^2 + q - 2r$ equals

• A. 8
• B. 16
• C. 32
• D. 24

Answer 1

Answer: (B)

16

Answer 2

Let the equation of the curve be $y = f(x)$. The curve passes through the origin, so $f(0) = 0$. Let $P(x, y)$ be a point on the curve. The equation of the tangent at $P(x, y)$ is $Y - y = \frac{dy}{dx}(X - x)$. Let $m = \frac{dy}{dx}$. The tangent is $Y - y = m(X - x)$. The tangent intersects the line $y=3$ at point $Q$. Let the coordinates of $Q$ be $(X_Q, 3)$. Substituting $Y=3$ in the tangent equation: $3 - y = m(X_Q - x)$. $X_Q - x = \frac{3 - y}{m} \implies X_Q = x + \frac{3 - y}{m}$. So, $Q = \left( x + \frac{3 - y}{m}, 3 \right)$.

The segment of the tangent drawn at $P(x, y)$ and the line $y=3$ is the segment PQ, where $P(x, y)$ and $Q\left( x + \frac{3 - y}{m}, 3 \right)$. The midpoint of the segment PQ is $M = \left( \frac{x + \left( x + \frac{3 - y}{m} \right)}{2}, \frac{y + 3}{2} \right) = \left( \frac{2x + \frac{3 - y}{m}}{2}, \frac{y + 3}{2} \right) = \left( x + \frac{3 - y}{2m}, \frac{y + 3}{2} \right)$.

The problem states that this midpoint M is bisected by the line $x + y = 1$. This means M lies on the line $x + y = 1$. Substituting the coordinates of M into the equation $x + y = 1$: $\left( x + \frac{3 - y}{2m} \right) + \left( \frac{y + 3}{2} \right) = 1$. Replace $m$ with $\frac{dy}{dx}$: $x + \frac{3 - y}{2 \frac{dy}{dx}} + \frac{y + 3}{2} = 1$. Multiply the entire equation by $2 \frac{dy}{dx}$: $2x \frac{dy}{dx} + (3 - y) + (y + 3) \frac{dy}{dx} = 2 \frac{dy}{dx}$. Rearrange the terms to form a differential equation: $(2x + y + 3 - 2) \frac{dy}{dx} = y - 3$. $(2x + y + 1) \frac{dy}{dx} = y - 3$. $\frac{dy}{dx} = \frac{y - 3}{2x + y + 1}$.

This is a first-order differential equation. To solve it, we can use a substitution to make it homogeneous. Let $y - 3 = v$ and $x = X - 2$. Then $dy = dv$ and $dx = dX$. $\frac{dv}{dX} = \frac{v}{2(X - 2) + (v + 3) + 1} = \frac{v}{2X - 4 + v + 4} = \frac{v}{2X + v}$. This is a homogeneous equation in $v$ and $X$. Let $v = uX$. Then $\frac{dv}{dX} = u + X \frac{du}{dX}$. $u + X \frac{du}{dX} = \frac{uX}{2X + uX} = \frac{u}{2 + u}$. $X \frac{du}{dX} = \frac{u}{2 + u} - u = \frac{u - u(2 + u)}{2 + u} = \frac{u - 2u - u^2}{2 + u} = \frac{-u - u^2}{2 + u} = - \frac{u(1 + u)}{2 + u}$. Separating variables: $\frac{2 + u}{u(1 + u)} du = - \frac{dX}{X}$. Using partial fraction decomposition for the left side: $\frac{2 + u}{u(1 + u)} = \frac{A}{u} + \frac{B}{1 + u}$. $2 + u = A(1 + u) + Bu$. Setting $u=0$, we get $2 = A(1) \implies A = 2$. Setting $u=-1$, we get $2 - 1 = B(-1) \implies 1 = -B \implies B = -1$. So, the equation is $\left( \frac{2}{u} - \frac{1}{1 + u} \right) du = - \frac{dX}{X}$. Integrating both sides: $\int \left( \frac{2}{u} - \frac{1}{1 + u} \right) du = \int - \frac{dX}{X}$. $2 \ln|u| - \ln|1 + u| = - \ln|X| + C_1$. $\ln\left|\frac{u^2}{1 + u}\right| = - \ln|X| + C_1$. $\ln\left|\frac{u^2}{1 + u}\right| + \ln|X| = C_1$. $\ln\left|\frac{u^2 X}{1 + u}\right| = C_1$. $\frac{u^2 X}{1 + u} = e^{C_1} = K$, where $K$ is a non-zero constant. (We need to consider the case $K=0$ later).

Substitute back $u = \frac{v}{X}$: $\frac{\left(\frac{v}{X}\right)^2 X}{1 + \frac{v}{X}} = K$. $\frac{\frac{v^2}{X}}{\frac{X + v}{X}} = K$. $\frac{v^2}{X + v} = K$. $v^2 = K(X + v)$.

Substitute back $v = y - 3$ and $X = x + 2$: $(y - 3)^2 = K((x + 2) + (y - 3))$. $(y - 3)^2 = K(x + 2 + y - 3)$. $(y - 3)^2 = K(x + y - 1)$.

The curve passes through the origin (0, 0). Substitute $x=0, y=0$ into the equation: $(0 - 3)^2 = K(0 + 0 - 1)$. $(-3)^2 = K(-1)$. $9 = -K \implies K = -9$.

Thus, the equation of the curve is $(y - 3)^2 = -9(x + y - 1)$. We are given the form $(y-p)^2 = q(r-x-y)$. $(y - 3)^2 = -9(x + y - 1)$. $(y - 3)^2 = 9(-(x + y - 1))$. $(y - 3)^2 = 9(1 - x - y)$. Comparing this with $(y-p)^2 = q(r-x-y)$, we have: $p = 3$. $q = 9$. $r = 1$. These are integers as stated in the question.

We need to calculate $p^2 + q - 2r$. $p^2 + q - 2r = (3)^2 + 9 - 2(1)$. $p^2 + q - 2r = 9 + 9 - 2$. $p^2 + q - 2r = 18 - 2$. $p^2 + q - 2r = 16$.

We should also consider the case where $K=0$. This would mean $v^2 = 0$, which implies $v = 0$. $y - 3 = 0 \implies y = 3$. The curve $y=3$ is a horizontal line. The tangent at any point $(x, 3)$ on this line is the line itself, $y=3$. The segment of the tangent drawn at $(x, 3)$ and the line $y=3$ is just the point $(x, 3)$. The midpoint of this segment (which is just the point itself) is $(x, 3)$. This midpoint must lie on the line $x+y=1$. So, $x + 3 = 1$, which means $x = -2$. This implies that the line $y=3$ satisfies the condition only at the point $(-2, 3)$. However, the condition must hold for any point $P(x, f(x))$ on the curve. So $y=3$ is not the required curve. Also, the curve passes through the origin (0,0), which is not on the line $y=3$. So $K=0$ is not possible.

The value of $p^2 + q - 2r$ is 16.