
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
160.8k+ views
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.
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.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

JEE Advanced 2025 Notes

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025
