Question:
If the curves $$ 2x^2 + ky^2 = 30 \quad \text{and} \quad 3y^2 = 28x $$ cut each other orthogonally, then \( k = \)
If the curves $$ 2x^2 + ky^2 = 30 \quad \text{and} \quad 3y^2 = 28x $$ cut each other orthogonally, then \( k = \)
Show Hint
For curves intersecting orthogonally, use the condition \( m_1 m_2 = -1 \), where \( m_1 \) and \( m_2 \) are the slopes of the tangents at the intersection point.
Updated On: Mar 18, 2026
- \( 5 \)
- \( 3 \)
- \( 2 \)
- \( 1 \)
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Find the Slopes of the Tangents
Differentiate the first equation: \[ 2x^2 + ky^2 = 30. \] \[ 4x + 2ky \frac{dy}{dx} = 0. \] \[ \frac{dy}{dx} = -\frac{2x}{ky}. \] Differentiate the second equation: \[ 3y^2 = 28x. \] \[ 6y \frac{dy}{dx} = 28. \] \[ \frac{dy}{dx} = \frac{14}{3y}. \] Step 2: Use the Orthogonality Condition
For curves to intersect orthogonally: \[ m_1 m_2 = -1. \] Substituting values: \[ \left(-\frac{2x}{ky} \right) \times \left( \frac{14}{3y} \right) = -1. \] \[ \frac{-28x}{3ky^2} = -1. \] \[ 28x = 3ky^2. \] Step 3: Solve for \( k \)
Using \( 3y^2 = 28x \): \[ k(28x) = 28x. \] \[ k = 1. \] Step 4: Conclusion
Thus, the correct answer is: \[ \mathbf{1}. \]
Was this answer helpful?
0
0
Top AP EAMCET Mathematics Questions
- If $\int \cos x . \cos 2x . \cos 5x dx = A \; \sin 2x + B \sin 4x + C \sin 6x + D \sin 8x + k $ (where $k$ is the arbitrary constant of integration), then $\frac{1}{B} + \frac{1}{C} = $
- If $k$ is one of the roots of the equation $x^2 - 25x + 24 = 0 $ such that $A = \begin{bmatrix}1&2&1\\ 3&2&3\\ 1&1&k\end{bmatrix} $ is a non-singular matrix, then $A^{-1}$ =
- If $p$ and $q$ are respectively the global maximum and global minimum of the function $f(x) = x^2 e^{2x}$ on the interval $[-2, 2]$ , then $pe^{-4} + qe^4 = $
- If $\displaystyle\sum^n_{k - 1} \tan^{-1} \left( \frac{1}{k^2 + k + 1} \right) = \tan^{-1} (\theta) $ , then $\theta$ =
- If the line joining the points $A(\alpha)$ and $B(\beta)$ on the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$ is a focal chord, then one possible values of $\cot \frac{\alpha}{2} . \cot \frac{\beta}{2}$ is
View More Questions
Top AP EAMCET Differentiation Questions
- If \( y = \sinh^{-1} \left(\frac{1 - x}{1 + x} \right) \), then \( \frac{dy}{dx} \) is given by:
- If \[ y = (x - 1)(x + 2)(x^2 + 5)(x^4 + 8), \] then \[ \lim\limits_{x \to -1} \left( \frac{dy}{dx} \right) = ? \]
- If \( y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \cdots \infty}}} \), then the value of \( \frac{d^2y}{dx^2} \) at \( (\pi,1) \) is:
- If \( f(0) = 0 \), \( f'(0) = 3 \), then the derivative of \( y = f(f(f(f(f(x))))) \) at \( x = 0 \) is:
- If \[ y = \tan^{-1} \left( \frac{2 - 3\sin x}{3 - 2\sin x} \right), \] then find \( \frac{dy}{dx} \).
View More Questions
Top AP EAMCET Questions
- If $\int \cos x . \cos 2x . \cos 5x dx = A \; \sin 2x + B \sin 4x + C \sin 6x + D \sin 8x + k $ (where $k$ is the arbitrary constant of integration), then $\frac{1}{B} + \frac{1}{C} = $
- If $k$ is one of the roots of the equation $x^2 - 25x + 24 = 0 $ such that $A = \begin{bmatrix}1&2&1\\ 3&2&3\\ 1&1&k\end{bmatrix} $ is a non-singular matrix, then $A^{-1}$ =
- If only $\frac{1}{51}^{th}$ ot the main current is to be passed through a galvanometer then the shunt required is $R_1$ and if only $\frac{1}{11}^{th}$ of the main voltage is to be developed across the galvanometer, then the resistance required is $R_2$. Then $\frac{R_2}{R_1} =$
- If $p$ and $q$ are respectively the global maximum and global minimum of the function $f(x) = x^2 e^{2x}$ on the interval $[-2, 2]$ , then $pe^{-4} + qe^4 = $
- If $\displaystyle\sum^n_{k - 1} \tan^{-1} \left( \frac{1}{k^2 + k + 1} \right) = \tan^{-1} (\theta) $ , then $\theta$ =
View More Questions