For what value of a do the curves y = 1 - ax^ | vedclass

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(B) Given curves are $y = 1 - ax^2$ and $y = x^2$.For the first curve,$\frac{dy}{dx} = -2ax$.For the second curve,$\frac{dy}{dx} = 2x$.Since the curve

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$1/3$

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(B) Given curves are $y = 1 - ax^2$ and $y = x^2$.For the first curve,$\frac{dy}{dx} = -2ax$.For the second curve,$\frac{dy}{dx} = 2x$.Since the curves intersect orthogonally,the product of their slopes at the point of intersection $(x, y)$ must be $-1$.So,$(-2ax)(2x) = -1 \implies 4ax^2 = 1 \implies x^2 = \frac{1}{4a}$.Now,equate the $y$ values of both curves at the intersection point:$1 - ax^2 = x^2 \implies 1 = (1 + a)x^2$.Substitute $x^2 = \frac{1}{4a}$ into the equation:$1 = (1 + a) \left( \frac{1}{4a} \right)$.$4a = 1 + a \implies 3a = 1 \implies a = 1/3$.

For what value of $a$ do the curves $y = 1 - ax^2$ and $y = x^2$ intersect orthogonally?

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