Question

Assertion :A normal is drawn at a point $P(x,y)$ of a curve. It meets the x-axis and the y-axis in point $A$ and $B$, respectively, such that $\frac{1}{OA}+\frac{1}{OB}=1$, where $O$ is the origin. The equation of such a curve passing through $(5,4)$ is ${(x-1)}^2+{(y-1)}^2=25$ Reason: $OA=x+y\frac{dy}{dx}$ and $OB=\frac{x+y\frac{dy}{dx}}{\frac{dy}{dx}}$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

The equation of the normal at $(x,y)$ is,

$(X-x)+(Y-y)\frac{dy}{dx}=0$
$x.e^y(\cos y-\sin y)=e^y\sin y+c$

$\Rightarrow \frac{X}{x+y\frac{dy}{dx}}+\frac{Y}{\frac{(x+y\frac{dy}{dx})}{\frac{dy}{dx}}}=1$

$\Rightarrow OA=x+y\frac{dy}{dx},OB=\frac{x+y\frac{dy}{dx}}{\frac{dy}{dx}}$
Also, $\frac{1}{OA}+\frac{1}{OB}=1\Rightarrow 1+\frac{dy}{dx}=x+y\frac{dy}{dx}\Rightarrow (y-1)\frac{dy}{dx}+(x-1)=0$
Integrating, we get
${(y-1)}^2+{(x-1)}^2=c$
Since the curve passes through $(5,4),c=25$
Hence, the curve is ${(x-1)}^2+{(y-1)}^2=25$