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  • Please solve RD Sharma class 12 chapter Differential Equations exercise 21.11 question 25 maths textbook solution

Please solve RD Sharma class 12 chapter Differential Equations exercise 21.11 question 25 maths textbook solution

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Answer: y^{2}=4 x

Given: Distance between foot of ordinate of the point of contact and the point of intersection of tangent and x \text {-axis }=2 x

To find: we have to find the equation of curve which passes through the point \left ( 1,2 \right )

Hint: Use equation of tangent at  P(x, y) \text { is }(Y-y)=\frac{d y}{d x}(X-x)

Solution: Equation of tangent at P(x,y)

Put Y=0

        \begin{aligned} &=-y=\frac{d y}{d x}(X-x) \\\\ &=-y \frac{d x}{d y}=X-x \\\\ &=X=x-y \frac{d x}{d y} \end{aligned}

Let co-ordinate of   B=\left(x-y \frac{d x}{d y}, 0\right)

Given distance between foot of ordinate of the point of contact and the point of intersection of tangent and \mathrm{x} \text {-axis }=2 x \text { i.e. } B C=2 x

        =\sqrt{\left(x-y \frac{d x}{d y}-x\right)^{2}+(0)^{2}}=2 x            [Distance formula  =\sqrt{\left(x-x_{1}\right)^{2}+\left(y-y_{1}\right)^{2}}]

        \begin{aligned} &=\sqrt{(-1)^{2}\left(y \frac{d x}{d y}\right)^{2}}=2 x \\\\ &=\sqrt{1\left(y \frac{d x}{d y}\right)^{2}}=2 x \end{aligned}

        \begin{aligned} &=y \frac{d x}{d y}=2 x \\\\ &=\frac{d x}{x}=2 \frac{d y}{y} \end{aligned}        [Separating variables]

Integration on both sides we get

        =\int \frac{dx}{x}=2 \int \frac{d y}{y}

        \begin{aligned} &=\log x=2 \log y+\log C \ldots(i) \end{aligned}

It passes through \left ( 1,2 \right )

        \begin{array}{ll} =\log 1=2 \log 2+\log C \\\\ =0=2 \log 2+\log C & {[\log 1=0]} \end{array}

        \begin{aligned} &=-2 \log 2=\log C \\\\ &=\log (2)^{-2}=\log C \quad\left[2 \log x=\log x^{2}\right] \\\\ &=\log \frac{1}{4}=\log C \end{aligned}

Using antilogarithm

        =C=\frac{1}{4}

Substituting the value of C in equation (i) we get

        \begin{aligned} &=\log x=2 \log y+\log \left(\frac{1}{4}\right) \\\\ &=\log x=\log y^{2}+\log \left(\frac{1}{4}\right) \end{aligned}

        \begin{aligned} &=\log x=\log \frac{y^{2}}{4} \\\\ &=x=\frac{y^{2}}{4} \\\\ &=y^{2}=4 x \end{aligned}

Hence y^{2}=4 x is the equation of the curve.

 

 

 

Posted by

infoexpert26

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