Question

The line $x+y=2$ is tangent to the curve $x^2=3-2y$ at its point

• A. $\left(1,1\right)$
• B. $\left(-1,1\right)$
• C. $\left(\sqrt{3},0\right)$
• D. $\left(3,-3\right)$

Answer 1

The correct option is A

$\left(1,1\right)$

Explanation for the correct option:

Given curve is $x^2=3-2y$
$\Rightarrow y=\frac{3-x^2}{2}$
The slope of a curve is obtained by differentiating it

Slope of the given curve is
$\begin{array}{rcl}& \Rightarrow & \frac{dy}{dx}=\frac{d}{dx}\left(\frac{3-x^2}{2}\right)\\ & \Rightarrow & \frac{dy}{dx}=-x\dots \left(1\right)\end{array}$

Since $x+y=2$ is tangent to $x^2=3-2y$ at the required point, their slopes will be equal at that point.

Slope of $x+y=2$ is the coefficient of $x$ in $y=mx+b$ form.
Converting the equation of the line into the above form,

$y=2-x$

Thus, the slope of the line is $-1$
Substituting this in $\left(1\right)$,

$$\begin{gathered} \Rightarrow -1=-x \\ \Rightarrow x=1 \end{gathered}$$

Putting the value of $x$ in $x+y=2$

$$\begin{gathered} \Rightarrow 1+y=2 \\ \Rightarrow y=1 \end{gathered}$$

Hence, point of intersection is $(1,1)$.

Option (A) is correct.