Question

Let a line $y=mx(m>0)$intersect the parabola,$y^2=x$ at a point $𝑃$, other than the origin. Let the tangent to it at $𝑃$ meet the $x-$axis at the point $Q$. If area$(\Delta 𝑂𝑃𝑄)=4sq$. units, then $m$ is equal to

Answer 1

Step 1: Illustrating the figure and Finding the equation of the tangent

Let $P(t^2,t)$ and be a point on the parabola then equation of tangent to the parabola at point

$\Rightarrow y-t=m_1\left(x-t^2\right)...\left(i\right)[\because \text{equation of line passing through}(x_1,y_1)isy-y_1=m_1\left(x-x_1\right)$

Where $m_1$ is slope of line and we know $m_1=\frac{dy}{dt}$

$$\begin{gathered} \because y^2=x\left[\text{given equation of parabola}\right] \\ \Rightarrow 2y\frac{dy}{dx}=1\left[\text{differentiating}w.r.t.x\right] \\ \Rightarrow \frac{dy}{dx}=\frac{1}{2y} \\ \Rightarrow m_1=\frac{1}{2t}\left[\because m_1=\frac{dy}{dx}atP\left(t^2,t\right)\Rightarrow y=t\right] \\ \Rightarrow y-t=\frac{1}{2t}\left(x-t^2\right)[from(i\left)\right] \\ \Rightarrow 2ty=x+t^2......\left(ii\right) \end{gathered}$$

$\because \left(ii\right)$ intersect the$x$ axis also so put $y=0$ in it

$$\begin{gathered} \Rightarrow 0=x+t^2 \\ \Rightarrow x=-t^2 \end{gathered}$$

So for triangle $OPQ$ coordinates of $Q$ be $\left(-t^2,0\right)$

Step 2: Finding Area of triangle :

The area of triangle having vertices $P(t^2,t)$ ,$Q\left(-t^2,0\right)$ and $O(0,0)$ $$\begin{gathered} \Rightarrow \frac{1}{2}\left|\begin{array}{ccc}0& 0& 1\\ t^2& t& 1\\ -t^2& 0& 1\end{array}\right|\left[\because \text{area of triangle having vertices}(p,q),(r,s)\&(t,u)=\frac{1}{2}\left|\begin{array}{ccc}p& q& 1\\ r& s& 1\\ t& u& 1\end{array}\right|\right] \\ \Rightarrow \frac{1}{2}\left|\begin{array}{ccc}0& 0& 1\\ t^2& t& 1\\ -t^2& 0& 1\end{array}\right|=4\left[\because \text{Area of}∆POQ=4given\right] \\ \Rightarrow 0-0+1\left(0-\left(-t^2·t\right)\right)=8 \\ \Rightarrow t^3=8 \\ \Rightarrow t=2...\left(iii\right)\left[\text{taking cube root both sides}\right] \end{gathered}$$

Step 3: Finding the value of m

The given equation of line $y=mx$, intersect the parabola $y^2=x$ then

$$\begin{gathered} \Rightarrow {\left(mx\right)}^2=x \\ \Rightarrow m^2=\frac{1}{x} \\ \Rightarrow m^2=\frac{1}{t^2}\left[\because \text{it passing through}Q(t^2,0)\right] \\ \Rightarrow m^{}=\frac{1}{t^{}}[\text{taking square root both sides}] \\ \Rightarrow m^{}=\frac{1}{2}[\text{from}(iii\left)\right] \end{gathered}$$

Thus the value of $m^{}=\frac{1}{2}$