Question

If the coordinates of two points $A$ and $B$ are $\left( {\sqrt 7 ,0} \right)$ and $\left( { - \sqrt 7 ,0} \right)$ respectively and $P$ is any point on the conic $9{x^2} + 16{y^2} = 144$, then $PA + PB$ is equal to
A. $6$
B. $16$
C. $9$
D. $8$

Answer 1

Hint: Find out the parametric coordinate of the point $P$. Then find the lengths of the line segments $A$ and $B$ and add them.

Formula Used:
Distance between the points $A\left( {{x_1},{y_1}} \right)$ and $B\left( {{x_2},{y_2}} \right)$ is $AB = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $ units.
The parametric coordinate of any point on the ellipse $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ is $\left( {a\cos \theta ,b\sin \theta } \right)$

Complete step by step solution:
The given equation of the conic is $9{x^2} + 16{y^2} = 144$, which represents an ellipse.
Convert the equation into the standard form of ellipse.
Divide both sides of the equation by $144$
$ \Rightarrow \dfrac{{9{x^2}}}{{144}} + \dfrac{{16{y^2}}}{{144}} = \dfrac{{144}}{{144}}\\ \Rightarrow \dfrac{{{x^2}}}{{16}} + \dfrac{{{y^2}}}{9} = 1$
Comparing this equation with the standard form of ellipse $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$, we get
${a^2} = 16$ and ${b^2} = 9$
${a^2} = 16 \Rightarrow a = \pm 4$
and ${b^2} = 9 \Rightarrow b = \pm 3$
The parametric coordinate of any point on the ellipse $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ is $\left( {a\cos \theta ,b\sin \theta } \right)$
In this problem, $a = \pm 4$ and $b = \pm 3$
Taking positive values, we get $a = 4$ and $b = 3$
So, the parametric coordinate is $\left( {4\cos \theta ,3\sin \theta } \right)$
Let this point be $P$.
Now, find the lengths of the line segments $PA$ and $PB$.
$PA = \sqrt {{{\left( {\sqrt 7 - 4\cos \theta } \right)}^2} + {{\left( {0 - 3\sin \theta } \right)}^2}} $
Use the expansion ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$
$ \Rightarrow \sqrt {7 - 8\sqrt 7 \cos \theta + 16{{\cos}^2}\theta + 9{{\sin}^2}\theta } $
Use the formula ${\sin ^2}\theta + {\cos ^2}\theta = 1$
$ \Rightarrow \sqrt {7 - 8\sqrt 7 \cos \theta + 16{{\cos}^2}\theta + 9\left( {1 - {{\cos}^2}\theta } \right)}$
$ \Rightarrow \sqrt {7 - 8\sqrt 7 \cos \theta + 16{{\cos}^2}\theta + 9 - 9{{\cos}^2}\theta } \\ = \sqrt {16 - 8\sqrt 7 \cos \theta + 7{{\cos}^2}\theta } \\ = \sqrt {{{\left( 4 \right)}^2} - 2 \times 4 \times \sqrt 7 \cos \theta + {{\left( {\sqrt 7 \cos \theta } \right)}^2}} \\ = \sqrt {{{\left( {4 - \sqrt 7 \cos \theta } \right)}^2}} \\ = 4 - \sqrt 7 \cos \theta $
$PB = \sqrt {{{\left( { - \sqrt 7 - 4\cos \theta } \right)}^2} + {{\left( {0 - 3\sin \theta } \right)}^2}} $
Use the expansion ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$
$ \Rightarrow \sqrt {7 + 8\sqrt 7 \cos \theta + 16{{\cos}^2}\theta + 9{{\sin}^2}\theta } $
Use the formula ${\sin ^2}\theta + {\cos ^2}\theta = 1$
$ \Rightarrow \sqrt {7 + 8\sqrt 7 \cos \theta + 16{{\cos}^2}\theta + 9\left( {1 - {{\cos}^2}\theta } \right)} \\ = \sqrt {7 + 8\sqrt 7 \cos \theta + 16{{\cos}^2}\theta + 9 - 9{{\cos}^2}\theta } \\ = \sqrt {16 + 8\sqrt 7 \cos \theta + 7{{\cos}^2}\theta } \\ = \sqrt {{{\left( 4 \right)}^2} + 2 \times 4 \times \sqrt 7 \cos \theta + {{\left( {\sqrt 7 \cos \theta } \right)}^2}} \\ = \sqrt {{{\left( {4 + \sqrt 7 \cos \theta } \right)}^2}} \\ = 4 + \sqrt 7 \cos \theta $
Add the lengths of the line segments $PA$ and $PB$.
$\therefore PA + PB = \left( {4 - \sqrt 7 \cos \theta } \right) + \left( {4 + \sqrt 7 \cos \theta } \right) = 8$

Option ‘D’ is correct

Note: Many students can’t remember the parametric coordinate of a point on an ellipse and the distance formula. You need to use the relation ${\sin ^2}\theta + {\cos ^2}\theta = 1$ to make the term under root a perfect square root.