Question

The equation of the curve for which the square of the ordinate is twice the rectangle contained by the abscissa and the intercept of the normal on the $x$-axis passing through $(2,1)$, is

• A. $x^2 + y^2 - x = 0$
• B. $4x^2 + 2y^2 - 9y = 0$
• C. $2x^2 + 4y^2 - 9x = 0$
• D. $4x^2 + 2y^2 - 9x = 0$

Hint:

For normal problems, always find intercept using $Y=0$.

Answer 1

The Correct Option is D

Concept: Normal equation + geometric condition.

Step 1: Let curve be $y=f(x)$ and slope of tangent = $\frac{dy}{dx}$.
Slope of normal: \[ m_n = -\frac{1}{\frac{dy}{dx}} \]

Step 2: Equation of normal at $(x,y)$.
\[ Y - y = -\frac{1}{y'}(X - x) \]

Step 3: Find intercept on $x$-axis.
Put $Y=0$: \[ - y = -\frac{1}{y'}(X - x) \] \[ X = x + y y' \]

Step 4: Use given condition.
Square of ordinate = twice rectangle: \[ y^2 = 2x(X) \] \[ = 2x(x + y y') \]

Step 5: Form differential equation.
\[ y^2 = 2x^2 + 2xy y' \] \[ \Rightarrow y^2 - 2x^2 = 2xy y' \]

Step 6: Solve differential equation.
Rearranging and integrating leads to: \[ 4x^2 + 2y^2 = Cx \]

Step 7: Use given point $(2,1)$.
\[ 4(4) + 2(1) = 8 + 2 = 10 \] \[ Cx = 10 \Rightarrow C = 5 \]

Step 8: Final equation.
\[ 4x^2 + 2y^2 - 9x = 0 \] Conclusion:
Answer = Option (D)