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  • If the curve ay + x^2 = 7 and x^3 = y, cut orthogonally at (1, 1), then the value of a is: A. 1 B. 0 C. – 6 D. 6

If the curve ay + x^2 = 7 and x^3 = y, cut orthogonally at (1, 1), then the value of a is:
A. 1
B. 0
C. – 6
D. 6

Answers (1)

Given the fact that curve ay + x^2 = 7 and x^3 = y, cut orthogonally at (1, 1)

Differentiate on both sides with x and get

\frac{\mathrm{d}\left(\mathrm{ay}+\mathrm{x}^{2}\right)}{\mathrm{dx}}=\frac{\mathrm{d}(7)}{\mathrm{dx}}

Apply sum rule and also 0 is the derivative of the constant, so
\Rightarrow \frac{\mathrm{d}(\mathrm{ay})}{\mathrm{dx}}+\frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{dx}}=0
Apply power rule and get
\\\Rightarrow \mathrm{a} \frac{\mathrm{dy}}{\mathrm{dx}}+2 \mathrm{x}=0\\\Rightarrow \frac{d y}{d x}=-\frac{2 x}{a}
Putting (1,1)

\\\Rightarrow\left(\frac{d y}{d x}\right)_{(1,1)}=-\frac{2 x}{a} \\\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(1,1)}=-\frac{2(1)}{\mathrm{a}}=-\frac{2}{\mathrm{a}} \ldots \ldots(\mathrm{i}) \\x^{3}=y
Differentiate on both sides with x and get
\frac{\mathrm{d}\left(\mathrm{x}^{3}\right)}{\mathrm{dx}}=\frac{\mathrm{d}(\mathrm{y})}{\mathrm{dx}}

Apply power rule and get
\Rightarrow 3 \mathrm{x}^{2}=\frac{\mathrm{dy}}{\mathrm{dx}}
Putting (1,1)
\\\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(1,1)}=3 \mathrm{x}^{2} \\\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(1,1)}=3(1)^{2}=3 \ldots \ldots(ii)

Both curves cut orthogonally at (1,1), 50
\left(\frac{d y}{d x}\right)_{(1,1)} \times\left(\frac{d y}{d x}\right)_{(1,1)}=-1
So from (i) and (ii), we get
\\\left(-\frac{2}{a}\right) \times 3=-1 \\\Rightarrow-\frac{6}{a}=-1\\\Rightarrow a=6

Hence when the curves cut orthogonally at (1, 1), then the value of a is 6.

So the correct answer is option D.

Posted by

infoexpert22

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