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 :
Answer & explanation
Correct answer: option 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.
$⇒f'(x)=2y\frac{dy}{dx}+3=0$
$⇒\frac{dy}{dx}=-\frac{3}{2y}$
$∴-\frac{3}{2k}=1⇒k=-\frac{3}{2}$