CAT Exam Model Paper 3 with solutions for free online practice

(d) Construct a function that describes the distance between ( x, y) and (4, 1).
By the “distance formula”, the distance, d, between ( x, y) and (4, 1) is
$\mathit{d}=\sqrt{{\left(\mathit{x}-4\right)}^2+{\left(\mathit{y}-1\right)}^2}$
For any point ( x, y) on the parabola, we have y = x${}^2$ +1,
Substituting for y in the formula for d, we have
$\mathit{d}=\sqrt{{\left(\mathit{x}-4\right)}^2+{\left(\left[{\mathit{x}}^2+1\right]-1\right)}^2}$
Simplifying, $\mathit{d}=\sqrt{{\left(\mathit{x}-4\right)}^2+{\left({\mathit{x}}^2\right)}^2}=\sqrt{{\left(\mathit{x}-4\right)}^2+{\mathit{x}}^4}$
Using the trace from function we can see that the minimum value of d occurs at (approximately) the point
x, d) = (1.12817, 3.4123)
That i.e., d is minimum when x is approximately 1.12817
The shortest distance between B and the parabola is approximately 3.14123
Using the equation for the parabola,y = x${}^2$ + 1 and the value of x that minimized d, ( x = 1.12817), we have y = 2.27277 (approximately).
So, the nearest point on the parabola y = x${}^2$ + 1, to the point (4, 1) is approximately (1.12817, 2.27277).