A curve y = f(x) passes through the...

MathematicsVariable Separable MethodJEE Advanced 1996Moderate

A curve $y = f (x)$ passes through the point $P (1, 1)$ . The normal to the curve at $P$ is $a (y - 1) + (x - 1) = 0$ . If the slope of the tangent at any point on the curve is proportional to the ordinate of the point, determine the equation of the curve. Also obtain the area bounded by the $y$ -axis, the curve and the normal to the curve at $P$ .

Answer:

y = e^{a (x - 1)}

; 1 sq. unit

Visualized Solution (Hindi)

Identify the Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P (1, 1)

Given point $P (1, 1)$ on the curve $y = f (x)$ .
The curve must satisfy the condition $f (1) = 1$ .

Analyze the Normal at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Equation of normal: $a (y - 1) + (x - 1) = 0$
Rearranging: $y - 1 = - \frac{1}{a} (x - 1)$
Slope of normal ( $m_{n}$ ) = $- \frac{1}{a}$

Find the Tangent Slope at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Slope of tangent ( $m_{t}$ ) $\times$ Slope of normal ( $m_{n}$ ) = $- 1$
Therefore, $m_{t} = - \frac{1}{m _{n}} = - \frac{1}{- 1/ a} = a$
At $P (1, 1)$ , $\frac{d y}{d x} = a$

Formulate the Differential Equation

Given: $\frac{d y}{d x} \propto y \Rightarrow \frac{d y}{d x} = k y$
At $(1, 1)$ , $\frac{d y}{d x} = a$ and $y = 1$
Substituting: $a = k (1) \Rightarrow k = a$
The differential equation is: $\frac{d y}{d x} = a y$

Solve by Variable Separation

Separate variables: $\frac{d y}{y} = a d x$
Integrate both sides: $\int \frac{d y}{y} = \int a d x$
$ln y = a x + C$

Find the Constant of Integration

Substitute $(1, 1)$ into $ln y = a x + C$ :
$ln (1) = a (1) + C \Rightarrow 0 = a + C \Rightarrow C = - a$
Equation: $ln y = a x - a = a (x - 1)$

Final Equation of the Curve

Taking exponential on both sides: $y = e^{a (x - 1)}$
This is the required equation of the curve.

Setup the Area Integral

Area bounded by $y$ -axis ( $x = 0$ ), curve $y = e^{a (x - 1)}$ , and normal $y = 1 - \frac{1}{a} (x - 1)$
Area $A = \int_{0}^{1} [y_{n or ma l} - y_{c u r v e}] d x$
$A = \int_{0}^{1} [(1 - \frac{1}{a} (x - 1)) - e^{a (x - 1)}] d x$

Evaluate the Integral

Integrating: $[x - \frac{1}{2 a} (x - 1)^{2} - \frac{1}{a} e^{a (x - 1)}]_{0}^{1}$
Upper limit ( $x = 1$ ): $1 - 0 - \frac{1}{a} = 1 - \frac{1}{a}$
Lower limit ( $x = 0$ ): $0 - \frac{1}{2 a} - \frac{e ^{- a}}{a}$
Area $A = (1 - \frac{1}{a}) - (- \frac{1}{2 a} - \frac{e ^{- a}}{a}) = 1 - \frac{1}{2 a} + \frac{e ^{- a}}{a}$

Conclusion and Summary

Final Curve: $y = e^{a (x - 1)}$
Calculated Area: $1 - \frac{1}{2 a} + \frac{e ^{- a}}{a}$ square units.
*Note: If $a \to \infty$ , the area approaches $1$ sq. unit.*

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Visualized Solution (Hindi)

Identify the Point $P (1, 1)$

Analyze the Normal at $P$

Find the Tangent Slope at $P$

Formulate the Differential Equation

Solve by Variable Separation

Find the Constant of Integration

Final Equation of the Curve

Setup the Area Integral

Evaluate the Integral

Conclusion and Summary

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Visualized Solution (Hindi)

Identify the Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(1,1)

Analyze the Normal at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Find the Tangent Slope at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Formulate the Differential Equation

Solve by Variable Separation

Find the Constant of Integration

Final Equation of the Curve

Setup the Area Integral

Evaluate the Integral

Conclusion and Summary

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