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### Answers (1)

Answer : $\left(x_{1}, y_{1}\right)=\left(\frac{1}{2}, 1\right)$

Hint :

For line, the slope of a line is denoted by m.

$y=m x+c$, where m is slope of line.

Given :

Given that the curve, $y^{2}=3-4 x$

Where the tangent parallel to the line, $2 x+y-2=0$

We have to find the co-ordinate of the point on the given curve.

Solution :

Let $\left(x_{1}, y_{1}\right)$ be the required point

We know that,

If  $2 x+y-2=0$

$\Rightarrow \quad y=-2 x+2$

Comparing with equation:

$y=mx+c$

We get, $m=-2$

Slope of the given line $=-2$

Since, the point lies on the curve

So         $y_{1}^{2}=3-4x_{1}$                                                                                      ...(i)

Now,      $y^{2}=3-4x$

Differentiate with respect to x

$\begin{array}{ll} \Rightarrow \quad & 2 y \frac{d y}{d x}=-4 \\ \therefore & \frac{d y}{d x}=\frac{-2}{y} \end{array}$

Slope of tangent $=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=\frac{-2}{y_{1}}$

Here,     slope of tangent=slope of the line

\begin{aligned} &\Rightarrow \quad \frac{-2}{y_{1}}=-2 \\ &\Rightarrow \quad y_{1}=1 \end{aligned}

From equation (i), we get

\begin{aligned} & 1=3-4 x_{1} \\ \Rightarrow & 4 x_{1}=2 \\ \Rightarrow & x_{1}=\frac{1}{2} \end{aligned}

Hence, $\left(x_{1}, y_{1}\right)=\left(\frac{1}{2}, 1\right)$

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