A curve C has the property that if the...

MathematicsVariable Separable MethodJEE Advanced 1998Moderate

A curve $C$ has the property that if the tangent drawn at any point $P$ on $C$ meets the co-ordinate axes at $A$ and $B$ , then $P$ is the mid-point of $A B$ . The curve passes through the point $(1, 1)$ . Determine the equation of the curve.

Answer:

x y = 1

Visualized Solution (English)

Visualizing the Problem

Let the curve be $C$ and a general point on it be $P (x, y)$ .
The tangent at $P$ meets the $x$ -axis at $A$ and the $y$ -axis at $B$ .
Given: $P$ is the midpoint of the segment $A B$ .
The curve passes through the point $(1, 1)$ .

Equation of the Tangent at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P (x, y)

The slope of the tangent at $P (x, y)$ is $m = \frac{d y}{d x} = y^{'}$ .
Equation of the tangent at $P (x, y)$ in point-slope form:
$Y - y = y^{'} (X - x)$

Finding Intercepts .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

For point $A$ ( $x$ -intercept), set $Y = 0$ :
$- y = y^{'} (X - x) \Rightarrow X = x - \frac{y}{y ^{'}}$
So, $A = (x - \frac{y}{y ^{'}}, 0)$
For point $B$ ( $y$ -intercept), set $X = 0$ :
$Y - y = y^{'} (0 - x) \Rightarrow Y = y - x y^{'}$
So, $B = (0, y - x y^{'})$

Applying the Midpoint Condition

Since $P (x, y)$ is the midpoint of $A B$ :
$x = \frac{( x - y / y ^{'} ) + 0}{2}$
$2 x = x - \frac{y}{y ^{'}}$
$x = - \frac{y}{y ^{'}}$

Forming the Differential Equation

Rearranging $x = - \frac{y}{y ^{'}}$ :
$y^{'} = - \frac{y}{x}$
Substituting $y^{'} = \frac{d y}{d x}$ :
$\frac{d y}{d x} = - \frac{y}{x}$

Solving by Variable Separation

Separate the variables $x$ and $y$ :
$\frac{1}{y} d y = - \frac{1}{x} d x$

Integrating Both Sides

Integrate both sides:
$\int \frac{1}{y} d y = - \int \frac{1}{x} d x$
$ln y = - ln x + ln k$
$ln y + ln x = ln k \Rightarrow ln (x y) = ln k$
$x y = k$

Applying the Initial Condition

The curve passes through $(1, 1)$ . Substitute $x = 1, y = 1$ :
$(1) (1) = k \Rightarrow k = 1$
The equation of the curve is $x y = 1$ .

Conclusion and Key Takeaway

Key Takeaway: The geometric property where the tangent midpoint lies on the curve leads to the differential equation $y^{'} = - y / x$ .
Final Equation: $x y = 1$ , which is a rectangular hyperbola.
Challenge: What if $P$ divided $A B$ in the ratio $1 : 2$ ?

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Comprehension Passage

A tangent

P T

is drawn to the circle

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at the point

P (3, 1)

. A straight line

L

, perpendicular to

P T

is a tangent to the circle

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A possible equation of $L$ is

(A)

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(B)

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(C)

x - 3 y = - 1

(D)

x + 3 y = 5

Answer:A

Question 2:

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(A)

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y = 2

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(D)

x + 22 y = 6

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