If the tangent to the curve $y^2+3x-7=0$ at the point (h, k) is parallel to the line $x-y =4, $ then the value of k is :

$-\frac{3}{2}$

The correct answer is Option (2) → $-\frac{3}{2}$

$x-y=4$

$⇒y=x-4$

∴ Slope = $m=1$   $[y=mx+c]$

∴ Slope of tangent to the curve must be 1.

$⇒f'(x)=2y\frac{dy}{dx}+3=0$

$⇒\frac{dy}{dx}=-\frac{3}{2y}$

$∴-\frac{3}{2k}=1⇒k=-\frac{3}{2}$