Question

Find the point on the straight line x+y=4, which is nearest to the parabola y²=4(x-9).

• A. (8,-4)
• B. (8,4)
• C. (-4,8)
• D. (4,8)

Answer 1

The correct option is A (8,-4)

The fig. shows a point $A$ on the parabola where the tangent to the parabola is parallel to the given line $x+y=4\cdots \left(1\right)$
It is clear that $A$ is the point on the parabola which is nearest to the given straight line. The slope of the tangent at any point on the parabola $y^2=4(x-9)\cdots \left(2\right)$
Is given by the equation $\frac{dy}{dx}=\frac{2}{y}\cdots \left(3\right)$

If $(h,k)$ be the coordinates of $A$, then $\frac{2}{k}=-1$ [tangent at $A$ is parallel to the given line $x+y=4]$

i.e. $k=-2$

and putting the value of $k$ in equation $\left(2\right)$, we have $h=\frac{k^2}{4}+9=10$
From the figure we see that if the normal drawn at $A$ intersects the given line at $B$, then $B$ is the required point on the line nearest to the parabola. Equation of the normal at $A$ is $y+2=1(x-10)$ [slope of normal at $A=1]$
i.e. $x-y=12\cdots \left(4\right)$
Solving equations $\left(1\right)$ and $\left(4\right)$ simultaneously gives the coordinates of $B$ as $(8,-4)$ which is the required point.