Question

A line L is perpendicular to the curve $y=\frac{x^2}{4}-2$ at its point P and passes through (10, -1). The coordinates of the point P are

• A. (2, -1)
• B. (6, 7)
• C. (0, -2)
• D. (4, 2)

Answer 1

The correct option is D (4, 2)

Let point $(x_1,y_1)$ on curve $y=\frac{x^2}{4}-2$
Slope of tangent is $\frac{dy}{dx}=\frac{x_1}{2}$
So slope of perpendicular line is $-\frac{2}{x_1}$ ...(1)
That perpendicular line also passes through $(x_1,y_1)$ and $(10,-1)$
Slope of line from these point is $\frac{y_1+1}{x_1-10}$ ...(2)
Equating (1) and (2) and solving we get,
$x_1=4$ and $y_1=2$
Thus point is $(4,2)$
Hence, option 'D' is correct.