Aspire Faculty ID #16719 · Topic: CUET 2023 · Just now
CUET 2023

If the curve $ay+x^2=7$ and $x^3=y$ cut orthogonally at $(1,1)$, then the value of $a$ is

Solution

From $ay+x^2=7$: $a\dfrac{dy}{dx}+2x=0$ $\Rightarrow \dfrac{dy}{dx}=-\dfrac{2x}{a}$ From $y=x^3$: $\dfrac{dy}{dx}=3x^2$ At $(1,1)$: $m_1=-\dfrac{2}{a},\quad m_2=3$ For orthogonal curves: $m_1m_2=-1$ $-\dfrac{2}{a}\times3=-1$ $\Rightarrow a=6$

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