Question

A and B are points $(-2,0)$ and $(1,3)$ on the curve $y=4-x^2$. If the tangent at P on the curve be parallel to chord AB, then co-ordinates of point P are

• A. $(-\frac{1}{3},\frac{5}{3})$
• B. $(\frac{1}{2},-\frac{15}{4})$
• C. $(-\frac{1}{2},\frac{15}{4})$
• D. $(-\frac{1}{3},\frac{1}{5})$

Answer 1

The correct option is C $(-\frac{1}{2},\frac{15}{4})$

Given equation of curve $y=4-x^2$ ...(i)
$\frac{dy}{dx}=-2x$
Given points $A(-2,0)$ and $(1,3)$
Slope of AB $=\frac{3}{3}=1$
$\Rightarrow -2x=1$
$\Rightarrow x=\frac{-1}{2}$
So, by equation (i), we get
$y=\frac{15}{4}$
Hence, the point is $(-\frac{1}{2},\frac{15}{4})$