Question

\(y=1-a x^{2}\) \(\frac{d y}{d x}=-2 a x_{1}\) \& \(y=x^{2}\) \(\frac{d y}{d x}=2 x_{1}\) If curves are orthogonal, \(\mathrm{m}_{1} \mathrm{~m}_{2}=-1\) \(+4 a x^{2}=1\) Now \(1-a x^{2}=x^{2} \Rightarrow \frac{1}{1+a}=x^{2}\) Using eq (1), we get \(\frac{4 a}{1+a}=1\) \(\mathrm{a}=\frac{1}{3}\)

Short answer

Answer: The value of 'a' for the curves to be orthogonal is \(\frac{1}{3}\).

Step-by-step solution

Finding the point of intersection

To find the point of intersection between the two curves, set the equations equal to each other: \(1 - ax^2 = x^2\) Solve for x: \(x^2 \cdot (1+a) = 1 \Rightarrow x^2 = \frac{1}{1+a}\)

Computing the derivatives

Compute the derivatives of both equations: 1. \(\frac{d(1-ax^2)}{dx} = -2ax\) 2. \(\frac{d(x^2)}{dx} = 2x\)

Checking if curves are orthogonal

To see if the curves are orthogonal at the point of intersection, their slopes should satisfy the condition \(m_1 \cdot m_2 = -1\). Substitute the derivatives and the point of intersection from Step 1: \((-2ax_1) \cdot (2x_1) = -1\) Substitute the value of \(x^2\) i.e \(\frac{1}{1+a}\) in the above equation and simplify: \(\frac{-4a}{1+a} = -1\)

Solving for 'a'

From Step 3, we have the equation: \(\frac{-4a}{1+a} = -1\) Solve for 'a': \(-4a = 1+a\) \(-a = \frac{1}{3}\) Hence, the value of 'a' for the curves to be orthogonal is \(\boxed{\frac{1}{3}}\).