Question

Let $P(x,y)$ be a point on the curve ${{y}^{2}}=4x$ at which the tangent is perpendicular to the line $2x+y=-2.$ Then, the coordinates of the point $P$ are

• A. $(4,\,4)$
• B. $(4,\,\,-\,4)$
• C. $(-4,\,\,4)$
• D. $(-4,\,\,-4)$

Answer 1

Answer: (A)

$(4,\,4)$

Answer 2

Given, ${{y}^{2}}=4x$
On differentiating w.r.t. x, we get
$2y\frac{dy}{dx}=4$
$\Rightarrow$ $\frac{dy}{dx}=\frac{2}{y}$
Since, the tangent to the curve is perpendicular to the line
$2x+y=-2$
$\therefore$ $\frac{2}{y}\times (-2)=-1$
$(\because \,\,\,{{m}_{1}}{{m}_{2}}=-1)$
$\Rightarrow$ $y=4$
$\therefore$ From E (i), ${{(4)}^{2}}=4x\,\Rightarrow x=4$
Hence, required point is $(4,4)$ .