Question

If the point P on the curve, $4{x^2} + 5{y^2} = 20$ is farthest from the point $Q(0, - 4)$, then find the value of $P{Q^2}$ .
A.48
B.29
C.21
D.36

Answer 1

Hint: Rewrite the given equation as the equation of an ellipse. Then take the parametric points of an ellipse and apply the distance formula to obtain the distance from the given point. Then put the maximum value of $\sin x$ to obtain the required result.

Formula Used:
The distance formula of two points $A(a,b),B(c,d)$ is,
$AB = \sqrt {{{(a - c)}^2} + {{(b - d)}^2}} $ .

Complete step by step solution:
The given equation can be written as $\dfrac{{{x^2}}}{{{{\left( {\sqrt 5 } \right)}^2}}} + \dfrac{{{y^2}}}{{{2^2}}} = 1$ ,
Hence the P can be taken as $\left( {\sqrt 5 \cos x + 2\sin x} \right)$ .
Now,
$P{Q^2} = {\left( {\sqrt 5 \cos x} \right)^2} + {\left( {2\sin x + 4} \right)^2}$
          $ = 5{\cos ^2}x + 4{\sin ^2}x + 16\sin x + 16$
          $ = 5({\cos ^2}x + {\sin ^2}x) - {\sin ^2}x + 16\sin x + 16$
          $ = - {\sin ^2}x + 16x + 21$
          $ = 85 - {(\sin x - 8)^2}$
The maximum value of $\sin x = 1$ .
Therefore, $P{Q^2} = 85 - 49 = 36$

Option ‘D’ is correct

Additional information
We can easily derive the ellipse equation when the ellipse's origin (0, 0) serves as its center and its x- and y-axes serve as its foci. The ellipse's equation is given as $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$.

Note: Sometimes, students simply let a point stand and use the formula for measuring distance to determine the necessary distance. But using that approach, they were unable to relate the given equation to anything, so we had to convert the equation first and take the parametric points to obtain the solution.